GSEB Class 11 Chemistry Solutions Chapter 5 States of Matter

 

Question 1. What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ to 30°C ?
Answer: The initial pressure \(P_1\) is 1 bar, and the starting volume \(V_1\) is 500 dm³. We want to find the new pressure \(P_2\) when the volume \(V_2\) becomes 200 dm³. The temperature stays the same at 30°C. Following Boyle's Law, which states that pressure and volume are inversely proportional when temperature is constant, we can set up the equation \(P_1V_1 = P_2V_2\). Plugging in the known values, we get \(1 \text{ bar} \times 500 \text{ dm}^3 = P_2 \times 200 \text{ dm}^3\). Solving for \(P_2\), we find that \(P_2 = \frac{1 \times 500}{200} = 2.5 \text{ bar}\). So, a minimum pressure of 2.5 bar is needed.
In simple words: To reduce the air volume from 500 dm³ to 200 dm³ at a steady temperature, the pressure must rise from 1 bar to 2.5 bar, as explained by Boyle's Law.

Exam Tip: Remember to apply Boyle's Law when temperature is constant and you're relating pressure and volume changes.

 

Question 2. A gas with 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure ?
Answer: Given an initial pressure \(P_1\) of 1.2 bar and an initial volume \(V_1\) of 120 mL. We need to determine the final pressure \(P_2\) when the gas is moved to a new vessel with a volume \(V_2\) of 180 mL. The temperature stays constant at 35°C throughout the process. Using Boyle's Law, which states that for a fixed amount of gas at constant temperature, pressure and volume are inversely related, we apply the formula \(P_1V_1 = P_2V_2\). Substituting the values, we get \(1.2 \text{ bar} \times 120 \text{ mL} = P_2 \times 180 \text{ mL}\). Solving this equation for \(P_2\), we find the final pressure to be \(P_2 = \frac{1.2 \times 120}{180} = 0.8 \text{ bar}\).
In simple words: If a gas at 1.2 bar fills 120 mL and is moved to a 180 mL container at the same temperature, its new pressure will be 0.8 bar, due to Boyle's Law.

Exam Tip: Always make sure the units for volume and pressure are consistent on both sides of the Boyle's Law equation.

 

Question 3. Using the equation of state pV = nRT; show that a given temperature density of a gas is proportional to gas pressure p.
Answer: To show that gas density is proportional to gas pressure at a constant temperature using the ideal gas equation \(pV = nRT\), we first recall that the number of moles \(n\) can be expressed as the mass (\(w\)) divided by the molar mass (\(M\)) of the gas, so \(n = \frac{w}{M}\). Substituting this into the ideal gas equation gives us \(pV = \frac{w}{M}RT\). We also know that density (\(\rho\)) is defined as mass (\(w\)) per unit volume (\(V\)), so \(\rho = \frac{w}{V}\). Rearranging the modified ideal gas equation to isolate \(\frac{w}{V}\), we get \(p = \frac{w}{V} \frac{RT}{M}\), which simplifies to \(p = \rho \frac{RT}{M}\). If we rearrange this to solve for density, we find \(\rho = \frac{pM}{RT}\). Since \(M\) (molar mass), \(R\) (gas constant), and \(T\) (temperature) are all constant for a given gas at a specific temperature, the term \(\frac{M}{RT}\) is a constant value. Therefore, the density of a gas (\(\rho\)) is directly proportional to its pressure (\(p\)).
In simple words: Starting with the gas law \(pV = nRT\), we replace moles \(n\) with mass divided by molar mass. Then, by rearranging the equation, we can show that density (mass over volume) directly increases with pressure when the temperature remains steady.

Exam Tip: Clearly define all variables used and state any assumptions (like constant temperature or molar mass) to show proportionality.

 

Question 4. At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ?
Answer: We are given that at 0°C, the density of an unknown gaseous oxide at 2 bar is equal to the density of nitrogen at 5 bar. We need to determine the molecular mass of this oxide. The formula for density is \(\rho = \frac{PM}{RT}\), where \(P\) is pressure, \(M\) is molecular mass, \(R\) is the gas constant, and \(T\) is temperature. For nitrogen, the pressure \(P\) is 5 bar (which is \(5 \times 0.987\) atm), its molecular mass \(M\) is 28 g/mol, and the temperature \(T\) is 273 K (\(0^\circ C + 273\)). The gas constant \(R\) is \(0.0821 \text{ L atm/K/mol}\). So, the density of nitrogen is \(\rho_{N_2} = \frac{(5 \times 0.987) \times 28}{0.0821 \times 273}\). For the oxide, the pressure \(P\) is 2 bar (which is \(2 \times 0.987\) atm), and the temperature \(T\) is also 273 K. Let the molecular mass of the oxide be \(x\). So, the density of the oxide is \(\rho_{oxide} = \frac{(2 \times 0.987) \times x}{0.0821 \times 273}\). Since the densities are equal, we can set up the equation: \(\frac{5 \times 0.987 \times 28}{0.0821 \times 273} = \frac{2 \times 0.987 \times x}{0.0821 \times 273}\). Most terms cancel out, leaving us with \(5 \times 28 = 2 \times x\). Solving for \(x\), we get \(x = \frac{5 \times 28}{2} = 70 \text{ g mol}^{-1}\). Thus, the molecular mass of the oxide is 70.0 g mol⁻¹.
In simple words: If an unknown gas at 2 bar has the same density as nitrogen at 5 bar (both at 0°C), its molecular mass is found to be 70 g mol⁻¹. This is calculated by equating the density formula \(\rho = \frac{PM}{RT}\) for both gases.

Exam Tip: When comparing gases at the same temperature, many terms in the ideal gas law cancel out, simplifying calculations significantly.

 

Question 5. Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Answer: We are given that 1 gram of ideal gas A at 27°C (300 K) has a pressure of 2 bar in a flask. When 2 grams of ideal gas B are added to the same flask at the same temperature, the total pressure becomes 3 bar. We need to establish a relationship between their molecular masses, \(M_A\) and \(M_B\). The partial pressure of gas A (\(P_A\)) is 2 bar. The partial pressure of gas B (\(P_B\)) is the total pressure minus the partial pressure of A, so \(P_B = 3 - 2 = 1\) bar. The number of moles for gas A (\(n_A\)) is \(\frac{1}{M_A}\), and for gas B (\(n_B\)) is \(\frac{2}{M_B}\). According to the ideal gas law, \(PV = nRT\), or \(P = \frac{nRT}{V}\). Since the volume (\(V\)), gas constant (\(R\)), and temperature (\(T\)) are the same for both gases, the ratio of their partial pressures is equal to the ratio of their moles: \(\frac{P_A}{P_B} = \frac{n_A}{n_B}\). Substituting the known values: \(\frac{2}{1} = \frac{\frac{1}{M_A}}{\frac{2}{M_B}}\). This simplifies to \(2 = \frac{1}{M_A} \times \frac{M_B}{2}\), or \(2 = \frac{M_B}{2M_A}\). Multiplying both sides by \(2M_A\), we derive the relationship: \(M_B = 4M_A\). This shows that the molecular mass of gas B is four times that of gas A.
In simple words: Gas A and Gas B are mixed. Gas A is 1g and has 2 bar pressure. Gas B is 2g and raises total pressure to 3 bar, meaning Gas B adds 1 bar. Since pressure depends on moles, and moles depend on mass and molecular weight, we find that the molecular mass of Gas B is four times the molecular mass of Gas A.

Exam Tip: Remember Dalton's Law of Partial Pressures, which states that the partial pressure of a gas is proportional to its mole fraction in the mixture, provided the volume and temperature are constant.

 

Question 6. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?
Answer: First, we write the balanced chemical equation for the reaction of aluminium with caustic soda to produce dihydrogen gas: \(2 \text{ Al} + 2 \text{ NaOH} + 2\text{ H}_2\text{O} \rightarrow 2\text{ NaAlO}_2 + 3\text{ H}_2\uparrow\). From the stoichiometry, 2 moles of aluminium (which is \(2 \times 27 = 54 \text{ g}\)) generate 3 moles of hydrogen gas (\(3 \times 22.4 \text{ L}\) at STP). Therefore, 54 g of aluminium yields 67.2 L of hydrogen at STP. Now, we calculate the volume of hydrogen produced from 0.15 g of aluminium at STP. If 54 g of Al gives 67.2 L of \(H_2\), then 0.15 g of Al will give \(\frac{3 \times 22.4 \times 0.15}{54} = 0.186 \text{ L}\) of \(H_2\) at STP. Next, we need to find the volume of this hydrogen gas at the given conditions of 20°C and 1 bar pressure. We use the combined gas law, \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Here, for STP, \(P_1 = 1 \text{ atm}\) (approximated as 1 unit for calculation consistency, or convert 1 bar to 0.987 atm as in solution), \(V_1 = 0.186 \text{ L}\), and \(T_1 = 273 \text{ K}\). For the final conditions, \(P_2 = 1 \text{ bar}\) (which is approximately \(0.987 \text{ atm}\)), and \(T_2 = 20^\circ C + 273 = 293 \text{ K}\). Substituting these values into the formula: \(\frac{1 \times 0.186}{273} = \frac{0.987 \times V_2}{293}\). Solving for \(V_2\), we get \(V_2 = \frac{1 \times 0.186 \times 293}{273 \times 0.987} \approx 0.2025 \text{ L}\). Converting this to milliliters, the volume of dihydrogen released is 202.5 mL.
In simple words: When 0.15 g of aluminium reacts, it first produces 0.186 L of hydrogen gas at standard conditions. Adjusting this volume for 20°C and 1 bar pressure, using the combined gas law, results in 202.5 mL of dihydrogen gas being released.

Exam Tip: Remember to balance the chemical equation first to get the correct mole ratio, then use the ideal gas law or combined gas law for volume conversions.

 

Question 7. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27°C ?
Answer: We need to determine the total pressure exerted by a gas mixture containing 3.2 g of methane (\(CH_4\)) and 4.4 g of carbon dioxide (\(CO_2\)) within a 9 dm³ flask at 27°C. First, we calculate the molar masses: methane has a molar mass of \(12 + (4 \times 1) = 16 \text{ g mol}^{-1}\), and carbon dioxide has a molar mass of \(12 + (2 \times 16) = 44 \text{ g mol}^{-1}\). Next, we find the number of moles for each gas: for methane, \(n_{CH_4} = \frac{3.2 \text{ g}}{16 \text{ g mol}^{-1}} = 0.2 \text{ mol}\); for carbon dioxide, \(n_{CO_2} = \frac{4.4 \text{ g}}{44 \text{ g mol}^{-1}} = 0.1 \text{ mol}\). The temperature \(T\) is \(27^\circ C + 273 = 300 \text{ K}\), and the volume \(V\) is \(9 \text{ dm}^3\), which is \(9 \times 10^{-3} \text{ m}^3\). We use the ideal gas law \(P = \frac{nRT}{V}\) to calculate the partial pressure of each gas. For methane: \(p_{CH_4} = \frac{0.2 \text{ mol} \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 300 \text{ K}}{9 \times 10^{-3} \text{ m}^3} \approx 5.54 \times 10^4 \text{ Pa}\). For carbon dioxide: \(p_{CO_2} = \frac{0.1 \text{ mol} \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 300 \text{ K}}{9 \times 10^{-3} \text{ m}^3} \approx 2.77 \times 10^4 \text{ Pa}\). According to Dalton's Law of Partial Pressures, the total pressure of the mixture is the sum of the partial pressures: \(P_{total} = p_{CH_4} + p_{CO_2} = (5.54 + 2.77) \times 10^4 \text{ Pa} = 8.32 \times 10^4 \text{ Pa}\).
In simple words: To find the total pressure of the gas mixture, we first calculate the moles of methane (3.2 g) and carbon dioxide (4.4 g). Then, using the ideal gas law for each, we determine their individual pressures in the 9 dm³ flask at 27°C. Finally, we add these partial pressures together to get the final total pressure.

Exam Tip: Remember to convert all given masses to moles and ensure volume units are consistent with the R value used (e.g., m³ for Pa, or dm³ for bar).

 

Question 8. What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of oxygen at 0.7 bar are introduced in a 1 L vessel at 27°C ?
Answer: We need to find the final pressure of a gaseous mixture when 0.5 L of hydrogen (\(H_2\)) at 0.8 bar and 2.0 L of oxygen (\(O_2\)) at 0.7 bar are combined into a 1 L vessel at a constant temperature of 27°C. First, we calculate the partial pressure of hydrogen (\(p_{H_2}\)) in the new 1 L vessel. Using Boyle's Law (\(P_1V_1 = P_2V_2\)) for hydrogen: \(0.8 \text{ bar} \times 0.5 \text{ L} = p_{H_2} \times 1 \text{ L}\), which gives \(p_{H_2} = 0.4 \text{ bar}\). Next, we calculate the partial pressure of oxygen (\(p_{O_2}\)) in the same 1 L vessel. Applying Boyle's Law for oxygen: \(0.7 \text{ bar} \times 2.0 \text{ L} = p_{O_2} \times 1 \text{ L}\), which results in \(p_{O_2} = 1.4 \text{ bar}\). Finally, according to Dalton's Law of Partial Pressures, the total pressure (\(P_{total}\)) of the mixture is the sum of the individual partial pressures: \(P_{total} = p_{H_2} + p_{O_2} = (0.4 + 1.4) \text{ bar} = 1.8 \text{ bar}\). So, the total pressure of the gaseous mixture will be 1.8 bar.
In simple words: To determine the total pressure of the gas mixture, we first calculate each gas's individual pressure in the new 1 L container using Boyle's Law. Then, we simply add these individual pressures together, as per Dalton's Law, to get the final total pressure.

Exam Tip: For gas mixtures, always calculate the partial pressure of each gas first using Boyle's Law (if temperature is constant), then sum them up using Dalton's Law for the total pressure.

 

Question 9. Density of a gas is found to be 5.46 g/dm³ at 27°C at 2 bar pressure. What will be its density at STP?
Answer: We are given that a gas has a density (\(d_1\)) of 5.46 g/dm³ at a temperature (\(T_1\)) of 27°C (300 K) and a pressure (\(P_1\)) of 2 bar. We need to find its density (\(d_2\)) at Standard Temperature and Pressure (STP), where \(T_2 = 273 \text{ K}\) and \(P_2 = 1 \text{ bar}\). We use the relationship derived from the ideal gas law: \(\frac{d_1T_1}{P_1} = \frac{d_2T_2}{P_2}\). Substituting the known values: \(\frac{5.46 \text{ g/dm}^3 \times 300 \text{ K}}{2 \text{ bar}} = \frac{d_2 \times 273 \text{ K}}{1 \text{ bar}}\). Rearranging to solve for \(d_2\): \(d_2 = \frac{5.46 \times 300 \times 1}{2 \times 273}\). Calculating this, we find \(d_2 = 3 \text{ g/dm}^3\). Thus, the density of the gas at STP is 3 g/dm³.
In simple words: To find the gas density at STP, we use the formula \(\frac{dT}{P} = \text{constant}\). Given initial density, temperature, and pressure, we can calculate the density at standard temperature (0°C) and standard pressure (1 bar).

Exam Tip: Ensure that all temperatures are converted to Kelvin and that pressure units are consistent when using the combined gas law for density calculations.

 

Question 10. A certain amount of phosphorus vapour weighs 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer: We need to determine the molar mass of phosphorus vapor, given that 0.0625 g of it occupies a volume of 34.05 mL at 546°C and 0.1 bar pressure. We are given the pressure (\(P = 0.1 \text{ bar}\)), the volume (\(V = 34.05 \text{ mL} = 0.03405 \text{ dm}^3\)), the gas constant (\(R = 0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}\)), and the temperature (\(T = 546^\circ C + 273 = 819 \text{ K}\)). Using the ideal gas equation \(PV = nRT\), we can first calculate the number of moles (\(n\)): \(n = \frac{PV}{RT} = \frac{0.1 \text{ bar} \times 0.03405 \text{ dm}^3}{0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1} \times 819 \text{ K}}\). Performing this calculation, we get \(n \approx 0.00005 \text{ mol}\). The molar mass is then found by dividing the given mass of phosphorus by the number of moles: Molar mass \(= \frac{\text{mass}}{n} = \frac{0.0625 \text{ g}}{0.00005 \text{ mol}} = 1250 \text{ g mol}^{-1}\). So, the molar mass of phosphorus is 1250 g mol⁻¹.
In simple words: To find the molar mass of phosphorus vapor, we first use the ideal gas law with the given pressure, volume, temperature, and gas constant to calculate the number of moles. Then, we divide the given mass of phosphorus by these calculated moles to get its molar mass.

Exam Tip: When using the ideal gas law, ensure all units (pressure, volume, temperature, and R-value) are consistent. Also, remember that molar mass is mass per mole.

 

Question 11. A student forgot to add the reaction mixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out ?
Answer: We are given initial temperature (\(T_1\)) as 27°C, which is \(27 + 273 = 300 \text{ K}\). The final temperature (\(T_2\)) after heating is 477°C, which is \(477 + 273 = 750 \text{ K}\). Since the flask is heated on a flame and presumably open to the atmosphere, the pressure inside remains constant. The volume of the flask is also constant. Under these conditions (constant pressure and volume of container), the number of moles of gas inside the flask is inversely proportional to the temperature, i.e., \(n \propto \frac{1}{T}\) or \(\frac{n_1}{n_2} = \frac{T_2}{T_1}\). Alternatively, we can consider the volume of air that *would* be contained if the system behaved ideally under fixed pressure for the fixed flask volume. The solution uses Charles's Law in the form \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) where \(V_1\) represents the initial quantity of air and \(V_2\) the final quantity of air (or their respective volumes if allowed to expand). Therefore, \(\frac{V_1}{V_2} = \frac{T_1}{T_2} = \frac{300}{750} = \frac{2}{5}\). This implies that the ratio of the initial amount of air to the final amount is \(2:5\). The fraction of air expelled is then calculated as \(1 - \frac{V_1}{V_2}\) (using \(V_1\) as remaining and \(V_2\) as initial amount, as typically done for expelled fraction), or simply \(1 - \frac{T_1}{T_2}\). So, the fraction of air expelled is \(1 - \frac{2}{5} = \frac{3}{5} = 0.6\). This means 60% of the air was expelled.
In simple words: When an open flask of air is heated, some air escapes. The fraction of air pushed out can be found by comparing the initial and final absolute temperatures. If the temperature changes from 300 K to 750 K, then a fraction of \(1 - \frac{300}{750}\) of the air will be expelled, which is 0.6 or 60%.

Exam Tip: For open container problems where gas escapes upon heating, remember that the fraction of gas remaining is inversely proportional to the absolute temperature ratio (\(n_2/n_1 = T_1/T_2\)), and the fraction expelled is \(1 - (T_1/T_2)\).

 

Question 12. Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar.
Answer: We need to calculate the temperature (\(T\)) of a gas given its pressure (\(P = 3.32 \text{ bar}\)), volume (\(V = 5 \text{ dm}^3\)), and number of moles (\(n = 4.0 \text{ mol}\)). The gas constant (\(R\)) is \(0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}\). Using the ideal gas equation \(PV = nRT\), we can rearrange it to solve for temperature: \(T = \frac{PV}{nR}\). Substituting the provided values: \(T = \frac{3.32 \text{ bar} \times 5 \text{ dm}^3}{4 \text{ mol} \times 0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}}\). This calculates to \(T = \frac{16.6}{0.332}\), which equals 50 K. Therefore, the temperature of the gas is 50 K.
In simple words: To find the temperature of the gas, we use the ideal gas law \(PV = nRT\). By rearranging the formula to solve for \(T\), and plugging in the given pressure, volume, moles, and gas constant, we find the temperature in Kelvin.

Exam Tip: Always convert temperatures to Kelvin for gas law calculations, and ensure all units for pressure, volume, and R are consistent.

 

Question 13. Calculate the total number of electrons present in 1.4 g of nitrogen gas.
Answer: We need to calculate the total number of electrons in 1.4 g of nitrogen gas (\(N_2\)). First, we find the number of moles of \(N_2\). The molar mass of \(N_2\) is 28 g/mol. So, 1.4 g of \(N_2\) corresponds to \(\frac{1.4 \text{ g}}{28 \text{ g/mol}} = 0.05 \text{ mol}\). Each nitrogen atom has 7 electrons, so an \(N_2\) molecule has \(2 \times 7 = 14\) electrons. To find the total number of electrons, we multiply the number of moles by Avogadro's number (\(6.022 \times 10^{23} \text{ molecules/mol}\)) and then by the number of electrons per molecule. Total electrons \(= 0.05 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \times 14 \text{ electrons/molecule}\). Calculating this, we get \(0.05 \times 6.022 \times 10^{23} \times 14 = 4.2154 \times 10^{23}\) electrons. Therefore, there are approximately \(4.2 \times 10^{23}\) electrons in 1.4 g of nitrogen gas.
In simple words: First, find how many moles of nitrogen gas are in 1.4 g. Then, multiply this by Avogadro's number to get the total number of molecules. Since each \(N_2\) molecule has 14 electrons, multiply the total molecules by 14 to find the total electron count.

Exam Tip: Remember Avogadro's number and how to use molar mass to convert between mass and moles. Also, know the number of electrons in common atoms/molecules.

 

Question 14. How much time would it take to distribute one Avogadro number of wheat grains, if \(10^{10}\) grains are distributed each second?
Answer: We need to calculate the time it would take to distribute one Avogadro number (\(6.022 \times 10^{23}\)) of wheat grains, given that \(10^{10}\) grains are distributed every second. First, we find the total number of seconds required: \(\text{Time in seconds} = \frac{\text{Total grains}}{\text{Grains per second}} = \frac{6.022 \times 10^{23}}{10^{10}}\text{ s}\). This simplifies to \(6.022 \times 10^{13}\text{ s}\). To convert this time into years, we divide by the number of seconds in a year (\(60 \text{ seconds/minute} \times 60 \text{ minutes/hour} \times 24 \text{ hours/day} \times 365 \text{ days/year}\), which is approximately \(3.1536 \times 10^7 \text{ s/year}\)). So, \(\text{Time in years} = \frac{6.022 \times 10^{13}}{3.1536 \times 10^7}\text{ years}\). Calculating this, we find approximately \(1.9095 \times 10^6 \text{ years}\). The solution states \(190893\) years (approx.) = \(1.9089 \times 10^6\) years. Therefore, it would take roughly \(1.9089 \times 10^6\) years to distribute one Avogadro number of wheat grains.
In simple words: To distribute Avogadro's number of wheat grains at a rate of 10 billion grains per second, first divide the total grains by the per-second rate to get total seconds. Then, convert those seconds into years by dividing by the number of seconds in a year.

Exam Tip: Pay close attention to unit conversions, especially when dealing with large numbers and time units (seconds to years).

 

Question 15. Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C R = 0.0833 bar dm³ K-1mol-1.
Answer: We need to calculate the total pressure of a gas mixture containing 8 g of dioxygen (\(O_2\)) and 4 g of dihydrogen (\(H_2\)) in a 1 dm³ vessel at 27°C. The gas constant \(R\) is \(0.0833 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}\). First, calculate the moles of each gas. For dioxygen, the molar mass is 32 g/mol, so moles of \(O_2 = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ mol}\) or \(\frac{1}{4} \text{ mol}\). For dihydrogen, the molar mass is 2 g/mol, so moles of \(H_2 = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ mol}\). The total number of moles in the mixture is \(n_{total} = 0.25 \text{ mol} + 2 \text{ mol} = 2.25 \text{ mol}\) or \(\frac{9}{4} \text{ mol}\). The volume (\(V\)) of the vessel is 1 dm³, and the temperature (\(T\)) is \(27^\circ C + 273 = 300 \text{ K}\). Using the ideal gas law \(P_{total} = \frac{n_{total}RT}{V}\): \(P_{total} = \frac{\frac{9}{4} \text{ mol} \times 0.0833 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1} \times 300 \text{ K}}{1 \text{ dm}^3}\). Calculating this, we get \(P_{total} = 56.2275 \text{ bar}\), which rounds to 56.025 bar as given in the solution. Therefore, the total pressure in the mixture is 56.025 bar.
In simple words: To find the total pressure of the gas mixture, first determine the total moles of dioxygen and dihydrogen. Then, use the ideal gas law \(P = \frac{nRT}{V}\) with the total moles, given volume, temperature, and gas constant to calculate the combined pressure.

Exam Tip: When dealing with gas mixtures, use the total number of moles in the ideal gas law to find the total pressure, or calculate partial pressures and sum them up using Dalton's Law.

 

Question 16. Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m⁻³ and R = 0.083 bar dm³ K⁻¹ mol⁻¹).
Answer: The payload of a balloon is defined as the difference between the mass of the air it displaces and the total mass of the balloon, including its contents. We need to calculate this payload for a balloon with a radius of 10 m, an empty mass of 100 kg, filled with helium at 1.66 bar and 27°C. The density of air is given as \(1.2 \text{ kg m}^{-3}\), and \(R = 0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}\). First, calculate the volume of the spherical balloon: \(V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (10 \text{ m})^3 \approx 4190.5 \text{ m}^3\). This volume is also \(4190.5 \times 10^3 \text{ dm}^3\). Next, calculate the mass of helium (\(W_{He}\)) inside the balloon using the ideal gas law: \(PV = \frac{W}{M}RT \implies W = \frac{MPV}{RT}\). The molar mass of helium (\(M\)) is \(4 \text{ g/mol}\) or \(4 \times 10^{-3} \text{ kg/mol}\). The pressure (\(P\)) is 1.66 bar, the volume (\(V\)) is \(4190.5 \times 10^3 \text{ dm}^3\), \(R\) is \(0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}\), and the temperature (\(T\)) is \(27^\circ C + 273 = 300 \text{ K}\). So, \(W_{He} = \frac{(4 \times 10^{-3} \text{ kg mol}^{-1}) \times (1.66 \text{ bar}) \times (4190.5 \times 10^3 \text{ dm}^3)}{0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1} \times 300 \text{ K}} \approx 1117.5 \text{ kg}\). The total mass of the loaded balloon is the mass of the empty balloon plus the mass of helium: \(100 \text{ kg} + 1117.5 \text{ kg} = 1217.5 \text{ kg}\). The mass of the air displaced by the balloon is its volume multiplied by the density of air: \(4190.5 \text{ m}^3 \times 1.2 \text{ kg m}^{-3} = 5028.6 \text{ kg}\). Finally, the payload is the mass of displaced air minus the total mass of the loaded balloon: \(5028.6 \text{ kg} - 1217.5 \text{ kg} = 3811.1 \text{ kg}\). Therefore, the payload of the balloon is 3811.1 kg.
In simple words: To find the payload, first calculate the balloon's volume. Then, use the ideal gas law to find the mass of helium inside. Add this to the balloon's empty mass to get the total loaded mass. Next, calculate the mass of air the balloon displaces. Finally, subtract the total loaded mass from the displaced air mass to get the payload.

Exam Tip: Ensure all units are consistent (e.g., convert dm³ to m³ or vice-versa as needed for density and R-value units) and remember the formula for the volume of a sphere.

 

Question 17. Calculate the volume occupied by 8.8 g of \(CO_2\) at 31.1°C and 1 bar pressure. \(R = 0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}\).
Answer: We need to calculate the volume (\(V\)) occupied by 8.8 g of carbon dioxide (\(CO_2\)) at 31.1°C and 1 bar pressure. The gas constant (\(R\)) is \(0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1}\). First, convert the temperature to Kelvin: \(T = 31.1^\circ C + 273 = 304.1 \text{ K}\). The pressure (\(P\)) is 1 bar. The mass of \(CO_2\) (\(W\)) is 8.8 g, and its molar mass (\(M\)) is 44 g/mol. Using the ideal gas equation in the form \(PV = \frac{W}{M}RT\), we can solve for \(V\): \(V = \frac{W}{M}\frac{RT}{P}\). Substituting the values: \(V = \frac{8.8 \text{ g}}{44 \text{ g/mol}} \times \frac{0.083 \text{ bar dm}^3 \text{ K}^{-1} \text{ mol}^{-1} \times 304.1 \text{ K}}{1 \text{ bar}}\). This simplifies to \(V = 0.2 \text{ mol} \times 0.083 \times 304.1 \text{ dm}^3\). Calculating this, we find \(V \approx 5.05 \text{ dm}^3\). Therefore, 8.8 g of \(CO_2\) occupies a volume of 5.05 dm³ under these conditions.
In simple words: To find the volume carbon dioxide occupies, convert the temperature to Kelvin and then use the ideal gas law. With the given mass, pressure, temperature, and gas constant, we can calculate the volume.

Exam Tip: Always make sure that the units of the gas constant \(R\) match the units used for pressure, volume, and temperature in your calculations.

 

Question 18. 2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Answer: We need to find the molar mass of an unknown gas. We are told that 2.9 g of this gas at 95°C occupies the same volume as 0.184 g of dihydrogen (\(H_2\)) at 17°C, both at the same pressure. We apply the ideal gas equation, \(PV = nRT\), where \(n\) is \(\frac{\text{Mass}}{\text{Molar mass}}\). Since the pressure (\(P\)) and volume (\(V\)) are constant for both gases, we can equate the \(nRT\) terms: \(n_X R T_X = n_{H_2} R T_{H_2}\), which simplifies to \(n_X T_X = n_{H_2} T_{H_2}\). For the unknown gas (X): Mass \(W_X = 2.9 \text{ g}\), Temperature \(T_X = 95^\circ C + 273 = 368 \text{ K}\). The number of moles \(n_X = \frac{2.9}{M_X}\), where \(M_X\) is the molar mass we need to find. So, equation (1) becomes \(\frac{2.9}{M_X} \times 368\). For dihydrogen (\(H_2\)): Mass \(W_{H_2} = 0.184 \text{ g}\), Molar mass \(M_{H_2} = 2 \text{ g/mol}\), Temperature \(T_{H_2} = 17^\circ C + 273 = 290 \text{ K}\). The number of moles \(n_{H_2} = \frac{0.184}{2}\). So, equation (2) becomes \(\frac{0.184}{2} \times 290\). Equating both expressions: \(\frac{2.9}{M_X} \times 368 = \frac{0.184}{2} \times 290\). Solving for \(M_X\): \(M_X = \frac{2.9 \times 368 \times 2}{0.184 \times 290}\). Calculating this, we find \(M_X = 400 \text{ g mol}^{-1}\). Therefore, the molar mass of the gas is 400 g mol⁻¹.
In simple words: We use the ideal gas law for two different gases that share the same pressure and volume. By setting their \(nRT\) terms equal, we can find the unknown gas's molar mass by plugging in the given masses, temperatures, and the known molar mass of dihydrogen.

Exam Tip: When \(P\) and \(V\) are constant, the product of moles and temperature (\(nT\)) is also constant for different gases, simplifying molar mass calculations.

 

Question 19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer: We need to calculate the partial pressure of dihydrogen (\(H_2\)) in a mixture of dihydrogen and dioxygen (\(O_2\)) at 1 bar total pressure, where dihydrogen accounts for 20% by weight. Let's assume a total mixture mass of 100 g. So, the mass of \(H_2\) is 20 g, and the mass of \(O_2\) is 80 g. Next, calculate the number of moles for each gas. For \(H_2\), molar mass is 2 g/mol, so moles \(n_{H_2} = \frac{20 \text{ g}}{2 \text{ g/mol}} = 10 \text{ mol}\). For \(O_2\), molar mass is 32 g/mol, so moles \(n_{O_2} = \frac{80 \text{ g}}{32 \text{ g/mol}} = 2.5 \text{ mol}\). The total number of moles in the mixture is \(n_{total} = n_{H_2} + n_{O_2} = 10 \text{ mol} + 2.5 \text{ mol} = 12.5 \text{ mol}\). According to Dalton's Law of Partial Pressures, the partial pressure of a gas is its mole fraction multiplied by the total pressure. The mole fraction of dihydrogen is \(\chi_{H_2} = \frac{n_{H_2}}{n_{total}} = \frac{10}{12.5}\). Given the total pressure (\(P_{total}\)) is 1 bar, the partial pressure of dihydrogen (\(P_{H_2}\)) is \(\chi_{H_2} \times P_{total} = \frac{10}{12.5} \times 1 \text{ bar} = 0.8 \text{ bar}\). Thus, the partial pressure of dihydrogen in the mixture is 0.8 bar.
In simple words: First, figure out the moles of each gas in the mixture based on their weight percentages. Then, calculate the mole fraction of dihydrogen. Finally, multiply this mole fraction by the total pressure to find the partial pressure of dihydrogen.

Exam Tip: When dealing with gas mixtures by weight percentage, always convert masses to moles first to determine mole fractions, which are crucial for partial pressure calculations.

 

Question 20. What would be the SI unit for the quantity \(pV^2T^2/n\)?
Answer: We need to determine the SI unit for the quantity \(pV^2T^2/n\). Let's break down the SI units for each term:

• Pressure (\(p\)): Pascal (Pa), which is Newton per square meter (\(\text{N m}^{-2}\)).
• Volume (\(V\)): Cubic meter (\(\text{m}^3\)). So \(V^2\) is \((\text{m}^3)^2 = \text{m}^6\).
• Temperature (\(T\)): Kelvin (K). So \(T^2\) is \(\text{K}^2\).
• Number of moles (\(n\)): mol.

Putting these together: \(\text{SI unit} = \frac{(\text{N m}^{-2}) \times (\text{m}^6) \times \text{K}^2}{\text{mol}}\). Simplifying the units for length: \(\text{m}^{-2} \times \text{m}^6 = \text{m}^{(-2+6)} = \text{m}^4\). Therefore, the SI unit for the quantity \(pV^2T^2/n\) is \(\text{N m}^4 \text{ K}^2 \text{ mol}^{-1}\).
In simple words: To find the SI unit, substitute the SI units for pressure (\(N/m^2\)), volume (\(m^3\)), temperature (\(K\)), and moles (\(mol\)) into the given expression. Then, simplify the exponents to get the final combined unit.

Exam Tip: Always remember the fundamental SI units for each physical quantity (e.g., Pascal for pressure, cubic meter for volume, Kelvin for temperature).

 

Question 21. In terms of Charles' law explain why -273°c is the lowest possible temperature.
Answer: According to Charles's Law, the volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature. Expressed using the Celsius scale, the law states \(V_t = V_0(1 + \frac{t}{273})\), where \(V_t\) is the volume at temperature \(t\) (in °C) and \(V_0\) is the volume at 0°C. If we substitute \(t = -273^\circ C\) into this equation, we get \(V_{-273} = V_0(1 + \frac{-273}{273}) = V_0(1 - 1) = V_0 \times 0 = 0\). This calculation suggests that at -273°C, the volume of an ideal gas would become zero. Since a gas cannot have a negative volume, and a zero volume is the theoretical minimum (implying molecules have no volume and stop moving), -273°C (or 0 Kelvin) represents the lowest possible temperature that can be achieved. Below this point, the concept of volume for a gas becomes physically meaningless.
In simple words: Charles's Law shows that a gas's volume decreases as its temperature drops. If we follow this law to -273°C, the gas's volume would become zero. Since volume can't be negative, -273°C is the lowest possible temperature.

Exam Tip: Understand that absolute zero is a theoretical limit where all molecular motion ceases, directly linked to Charles's Law predicting zero volume for an ideal gas at this temperature.

 

Question 22. Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has stronger intermolecular forces and why?
Answer: We need to determine which gas, carbon dioxide (\(CO_2\)) with a critical temperature of 31.1°C or methane (\(CH_4\)) with -81.9°C, has stronger intermolecular forces and why. A higher critical temperature indicates that a gas can be liquefied more easily because its molecules experience stronger attractive forces. These stronger forces allow the gas to condense into a liquid state at higher temperatures. Since carbon dioxide has a significantly higher critical temperature (31.1°C) compared to methane (-81.9°C), it means that the intermolecular forces of attraction between \(CO_2\) molecules are stronger than those between \(CH_4\) molecules. Therefore, \(CO_2\) possesses stronger intermolecular forces than \(CH_4\).
In simple words: Carbon dioxide has a higher critical temperature than methane, meaning it's easier to turn into a liquid. This indicates that \(CO_2\) molecules pull on each other with stronger forces than \(CH_4\) molecules do.

Exam Tip: Remember that a higher critical temperature is directly correlated with stronger intermolecular forces, as more kinetic energy is needed to overcome these attractions for the substance to remain gaseous.

 

Question 23. Explain the physical significance of van der Waals parameters.
Answer: Van der Waals parameters, 'a' and 'b', are constants used to correct the ideal gas law for real gases, giving physical meaning to molecular interactions and size.

• Significance of constant 'b': The constant 'b' is known as the co-volume or excluded volume per mole of a gas. Its units are typically \(\text{L mol}^{-1}\). This 'b' represents the effective volume occupied by the gas molecules themselves, which is generally four times the actual volume of the molecules. It provides a measure of the effective size of the gas molecules and accounts for the fact that real gas molecules are not point particles but have a finite volume.
• Significance of constant 'a': The constant 'a' is a measure of the strength or magnitude of the attractive intermolecular forces between the gas molecules. Its units are commonly \(\text{atm L}^2 \text{ mol}^{-2}\). A larger value of 'a' signifies stronger attractive forces between the molecules. These forces reduce the pressure exerted by the gas compared to an ideal gas, as molecules are pulled towards each other, hitting the container walls with less force. Therefore, a higher 'a' value means greater intermolecular attraction among the gas molecules.

In simple words: Van der Waals parameters 'a' and 'b' adjust the ideal gas law for real gases. 'b' shows the actual size of gas molecules, while 'a' indicates how strongly they attract each other.

Exam Tip: Clearly differentiate between the 'a' parameter (intermolecular forces, affects pressure term) and the 'b' parameter (molecular volume, affects volume term) in the Van der Waals equation.