CBSE Class 11 Chemistry Thermodynamics MCQs Set 06

Answer: (d) Heat

Explanation:
1. A state function depends only on the initial and final states of the system.
2. Internal energy, enthalpy and entropy are state functions.
3. Heat depends on the path followed, so it is a path function.

Teacher's Note:
1. Remember that q and w are path functions.
2. Their sum ΔU = q + w is a state function.

Answer: (b) Work

Explanation:
1. Temperature, entropy and enthalpy are fixed by the state of the system.
2. Work done depends on how the change is carried out (for example reversibly or against a fixed pressure), so it is a path function.

Teacher's Note:
1. Ask: does the value change if the same initial and final states are joined by a different path?
2. If yes, it is a path function.

Answer: (c) Entropy

Explanation:
1. Intensive properties do not depend on the amount of matter.
2. Pressure, temperature and molar volume are the same for any portion of a uniform sample.
3. Entropy doubles when the amount of substance is doubled, so it is extensive.

Teacher's Note:
1. A property per mole or per gram is usually intensive.
2. Mass, volume, U, H, S and C are extensive.

Answer: (a) Internal energy

Explanation:
1. Internal energy depends only on the state and is proportional to the amount of substance.
2. Pressure, molar heat capacity and temperature are state properties but intensive.

Teacher's Note:
1. Check both conditions separately.
2. Enthalpy and entropy also satisfy both.

Answer: (c) an open system

Explanation:
1. The test tube is open, so the NH₃ gas formed leaves and heat is exchanged with the surroundings.
2. Both matter and energy cross the boundary, which defines an open system.

Teacher's Note:
1. Open: matter and energy exchanged. Closed: only energy. Isolated: neither.
2. Look at whether the vessel is sealed.

Answer: (d) an isolated system

Explanation:
1. A tight stopper prevents the exchange of matter.
2. The vacuum jacket and the walls stop heat flow, so no energy is exchanged either.
3. Neither matter nor energy is exchanged, so the system is isolated.

Teacher's Note:
1. A thermos flask is the standard practical example of an isolated system.
2. A sealed ordinary bottle would be a closed system.

Answer: (a) equal to zero

Explanation:
1. The gas returns to its initial state, so the process is cyclic.
2. U is a state function, so ΔU = U_final − U_initial = 0 for the complete cycle.

Teacher's Note:
1. For any cyclic process ΔU = 0 and ΔH = 0.
2. Then q = −w.

Answer: (b) The system gains heat and has work performed on it

Explanation:
1. ΔU = q + w. Gaining heat makes q positive and having work performed on the system makes w positive.
2. Both terms are positive, so ΔU is certainly positive.
3. In the other cases at least one term is negative, so the sign of ΔU is uncertain or negative.

Teacher's Note:
1. Use the IUPAC convention: energy entering the system is positive.
2. Decide the sign of q and w separately.

Answer: (c) ΔU = +15 kJ, so the internal energy increases

Explanation:
1. q is negative, so heat flows out of the system to the surroundings.
2. w is positive, so work is done on the system by the surroundings.
3. ΔU = q + w = −10 + 25 = +15 kJ, so the internal energy increases.

Teacher's Note:
1. Under the IUPAC convention positive w means work done on the system.
2. Add the signed values of q and w.

Answer: (d) 5 P₁V₁

Explanation:
1. The net work in a cycle equals the area enclosed by the loop; the cycle is clockwise, so the gas does net work.
2. The enclosed figure is a right-angled triangle with base AC = 3V₁ − V₁ = 2V₁ and height AB = 6P₁ − P₁ = 5P₁.
3. Area = ½ × 2V₁ × 5P₁ = 5P₁V₁.

Teacher's Note:
1. For a cyclic process ΔU = 0, so the work done by the gas equals the heat absorbed.
2. Read the coordinates of A, B and C carefully before computing the area.

Answer: (b) 10 kJ

Explanation:
1. At constant volume w = −pΔV = 0.
2. ΔU = q + w = q_V = 10 kJ.

Teacher's Note:
1. The heat absorbed at constant volume equals ΔU.
2. The temperature change is only extra data.

Answer: (c) 10 J

Explanation:
1. For an isothermal change of an ideal gas ΔU = 0.
2. So q + w = 0, giving w = −10 J, which means 10 J of work is done by the gas.

Teacher's Note:
1. Ideal-gas energy depends on temperature alone.
2. Heat absorbed is converted completely into work in this process.

Answer: (a) The energy of the universe remains constant

Explanation:
1. The first law is the law of conservation of energy.
2. Energy can flow between a system and its surroundings, so neither one alone stays constant, but the total energy of the universe is constant.

Teacher's Note:
1. Do not confuse this with the second law, which deals with entropy.
2. Entropy of the universe tends to increase.

Answer: (d) −83.48 J

Explanation:
1. p = 1520 torr = 2 atm and ΔV = 177 − 377 = −200 mL = −0.200 L.
2. w = −pΔV = −2 × (−0.200) = +0.400 L atm = +40.52 J (work done on the gas).
3. q = −124 J, so ΔU = q + w = −124 + 40.52 = −83.48 J.

Teacher's Note:
1. Convert torr to atm (760 torr = 1 atm) and mL to L before using 101.3 J per L atm.
2. Compression gives positive work.

Answer: (b) w > 0; q = 0; ΔU > 0

Explanation:
1. The stirrer does work on the liquid, so w > 0.
2. The calorimeter is insulated, so q = 0.
3. ΔU = q + w = w > 0, so the temperature of the liquid rises.

Teacher's Note:
1. Insulated means adiabatic (q = 0).
2. This is the idea behind Joule's paddle-wheel experiment.

Answer: (a) −47 kJ

Explanation:
1. w = −p_exΔV = −1.2 × 32.0 = −38.4 L atm = −3.89 kJ.
2. q = ΔU − w = −51.0 − (−3.89) = −47.1 kJ, about −47 kJ.

Teacher's Note:
1. Rearrange ΔU = q + w to q = ΔU − w.
2. The system gives out heat here.

Answer: (c) ΔU = 0

Explanation:
1. The internal energy of an ideal gas depends only on temperature.
2. The temperature is constant, so ΔU = 0.
3. The volume changes and the gas does work, and heat flows in to compensate (q = −w).

Teacher's Note:
1. Isothermal means constant temperature, not zero heat.
2. Zero heat is adiabatic.

Answer: (d) −1381.8 cal

Explanation:
1. w = −2.303 nRT log (V_f/V_i) with T = 300 K.
2. log (22.4/2.24) = log 10 = 1.
3. w = −2.303 × 1 × 2 × 300 × 1 = −1381.8 cal.

Teacher's Note:
1. The volume ratio is exactly 10, so the log is 1.
2. The negative sign means work is done by the gas.

Answer: (b) −11.488 kJ

Explanation:
1. w = −2.303 nRT log (V_f/V_i) = −2.303 × 2 × 8.314 × 300 × log 10.
2. = −11 488 J = −11.488 kJ.

Teacher's Note:
1. The ratio 40/4 = 10 makes the logarithm 1.
2. Convert J to kJ at the end.

Answer: (a) −11.224 kJ

Explanation:
1. w = −p_ex(V₂ − V₁) with V = nRT/p.
2. V₂ − V₁ = nRT(1/1 − 1/10) = 0.9 nRT (in atm⁻¹ units), and p_ex = 1 atm.
3. w = −0.9 × 5 × 8.314 × 300 = −11 224 J = −11.224 kJ.

Teacher's Note:
1. Irreversible work uses the external pressure, not the gas pressure.
2. The result is smaller in magnitude than the reversible work.