CBSE Class 10 Physics The Gas Laws MCQs Set 18

Answer: (d) temperature and pressure

Explanation:
1. A gas has no fixed volume, so its volume changes with the pressure on it and with its temperature.
2. Mass is fixed in this case and density depends on volume.

Teacher's Note:
1. Solids and liquids are almost incompressible, so their volume depends on temperature only.
2. To study one factor, the other is kept constant.

Answer: (c) solids expand least and gases expand most

Explanation:
1. Solids have strong forces between their particles and expand least.
2. Liquids expand more than solids.
3. Gases expand most.

Teacher's Note:
1. This is why gases are used in gas thermometers.
2. The coefficient of cubical expansion of a gas is about 1/273 per °C.

Answer: (c) 263 K

Explanation:
1. T = t + 273.
2. T = -10 + 273 = 263 K.

Teacher's Note:
1. The Kelvin scale never has negative values.
2. 0 K corresponds to -273 °C.

Answer: (c) zero

Explanation:
1. V is proportional to T.
2. At T = 0 K, V = 0, in theory.
3. Real gases liquefy before this temperature is reached.

Teacher's Note:
1. This is why absolute zero is called the ideal gas scale zero.
2. 0 K has never been reached exactly.

Answer: (d) 1/273

Explanation:
1. By Charles' law all gases expand by 1/273 of their volume at 0 °C for every °C rise, at constant pressure.
2. So the volume coefficient is 1/273 = 0.00366 per °C.

Teacher's Note:
1. The pressure coefficient at constant volume has the same value for a perfect gas.
2. The value is the same for all gases.

Answer: (a) a constant for a given mass of gas

Explanation:
1. Combining Boyle's and Charles' laws gives PV/T = constant.
2. Hence P1V1/T1 = P2V2/T2.

Teacher's Note:
1. This is called the general gas equation.
2. For one mole the constant is R = 8.31 J/mol K.

Answer: (b) Charles' law

Explanation:
1. Keeping the pressure fixed and varying only the temperature leads to V proportional to T.
2. That is Charles' law.

Teacher's Note:
1. Boyle's law is about P and V at constant temperature.
2. V1/T1 = V2/T2 for constant pressure.

Answer: (d) the product A × B

Explanation:
1. A is proportional to 1/B means A = k/B.
2. Hence AB = k, a constant.

Teacher's Note:
1. As B increases, A decreases.
2. For Boyle's law PV = constant at fixed temperature.

Answer: (c) both Boyle's law and Charles' law

Explanation:
1. At constant T it gives P1V1 = P2V2, which is Boyle's law.
2. At constant P it gives V1/T1 = V2/T2, which is Charles' law.
3. At constant V it gives P1/T1 = P2/T2, the pressure law.

Teacher's Note:
1. It is called the general gas equation.
2. It is used when pressure, volume and temperature all change.

Answer: (a) equal to it

Explanation:
1. Both coefficients equal 1/273 per °C for a perfect gas.
2. Hence they are equal.

Teacher's Note:
1. This is why the same absolute scale can be defined by either the constant-volume or the constant-pressure gas thermometer.
2. Real gases show small differences.

Answer: (c) 8.31 J mol⁻¹ K⁻¹

Explanation:
1. At S.T.P. one mole occupies 22.4 × 10⁻³ m³, P = 1.013 × 10⁵ Pa, T = 273 K.
2. R = PV/T = (1.013 × 10⁵ × 22.4 × 10⁻³) / 273 = 8.31 J/K mol.

Teacher's Note:
1. PV = RT for one mole and PV = nRT for n moles.
2. 1.013 × 10⁵ Pa is standard atmospheric pressure.

Answer: (b) 109 cm³

Explanation:
1. Absolute temperatures: T1 = 300 K, T2 = 327 K.
2. V2 = V1 T2 / T1 = 100 × 327 / 300 = 109 cm³.
3. The volume does not double when the Celsius temperature doubles.

Teacher's Note:
1. Always convert to kelvin before using Charles' law.
2. Volume doubles when the absolute temperature doubles, that is, from 27 °C to 327 °C.

Answer: (d) both pressure and temperature

Explanation:
1. Boyle's law shows V depends on P at constant T.
2. Charles' law shows V depends on T at constant P.
3. So it depends on both.

Teacher's Note:
1. The general gas equation combines the two.
2. The volume of a solid depends on temperature only.

Answer: (c) 303 K

Explanation:
1. T = 30 + 273 = 303 K.

Teacher's Note:
1. Add 273 to the Celsius temperature.
2. The size of one kelvin equals the size of one degree Celsius.

Answer: (c) Charles' law

Explanation:
1. This describes expansion with temperature at constant pressure.
2. That is Charles' law, also stated as V1/T1 = V2/T2.

Teacher's Note:
1. 1/273 is the volume coefficient.
2. Boyle's law concerns pressure and volume at constant temperature.

Answer: (b) 200 cm³

Explanation:
1. Boyle's law: P1V1 = P2V2.
2. P2 = 3P1, so V2 = 600 / 3 = 200 cm³.

Teacher's Note:
1. Volume falls to one third when the pressure triples.
2. The pressure and volume of a gas at fixed temperature are inversely proportional.

Answer: (a) 250 cm³

Explanation:
1. T1 = 47 + 273 = 320 K and T2 = 127 + 273 = 400 K.
2. V2 = V1 T2 / T1 = 200 × 400 / 320 = 250 cm³.

Teacher's Note:
1. Use kelvin temperatures.
2. Doubling the Celsius temperature would give the wrong answer.

Answer: (b) 96 cm of mercury

Explanation:
1. T1 = 300 K, T2 = 360 K and V is constant.
2. P2 = P1 T2 / T1 = 80 × 360 / 300 = 96 cm of mercury.

Teacher's Note:
1. At constant volume pressure is proportional to the absolute temperature.
2. This is Charles' pressure law.

Answer: (c) 1/273 per °C

Explanation:
1. V = V0(1 + α t).
2. 500 = V0(1 + 27α) and 600 = V0(1 + 87α).
3. Dividing, 5(1 + 87α) = 6(1 + 27α), so 273α = 1 and α = 1/273.

Teacher's Note:
1. The method works for any pair of volumes at the same pressure.
2. The value 1/273 shows the gas behaves ideally.

Answer: (a) 1/273 per °C

Explanation:
1. P = P0(1 + α t).
2. 72 = P0(1 + 27α) and 108 = P0(1 + 177α).
3. Dividing, 2(1 + 177α) = 3(1 + 27α), so 273α = 1.

Teacher's Note:
1. Pressure coefficient and volume coefficient are both 1/273 for a perfect gas.
2. The same working is used for both.

Answer: (a) 262.5 cm³

Explanation:
1. Boyle's law: P1V1 = P2V2.
2. 70 × 300 = 80 × V2, so V2 = 21 000 / 80 = 262.5 cm³.

Teacher's Note:
1. Pressure and volume change in opposite directions.
2. Check that V2 is smaller than V1 when pressure rises.

Answer: (b) 1.013 × 10⁵ Pa

Explanation:
1. P = hdg with h = 0.76 m.
2. P = 0.76 × 13.6 × 10³ × 9.8 = 1.013 × 10⁵ N/m² (Pa).

Teacher's Note:
1. 1 bar = 10⁵ Pa, so this is about 1.013 bar.
2. Convert h to metres.