physics : class11 Adiabatic Process and Internal Energy : Internal Energy : Thermodynamics MCQ Questions and Answers
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Adiabatic Process and Internal Energy
MCQ questions and answers.
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MCQ Questions
1. One mole of a monatomic ideal gas initially at 300 K undergoes a reversible adiabatic compression until its volume becomes half its initial value. What is the approximate increase in its internal energy? Take R = 8.31 J mol⁻¹ K⁻¹.
A.
1.10 × 10³ J
B.
2.20 × 10³ J
C.
3.32 × 10³ J
D.
4.40 × 10³ J
Correct Answer:
2.20 × 10³ J
Explanation:
For a monatomic ideal gas, γ = 5/3 and TV^(γ−1) is constant during a reversible adiabatic process. Thus T₂ = 300(2)^(2/3) ≈ 476 K. Therefore, ΔU = nCᵥ(T₂ − T₁) = (3/2)(8.31)(476 − 300) ≈ 2.20 × 10³ J.
2. An ideal gas undergoes a free expansion into an evacuated chamber inside a thermally insulated container. Which statement about its internal energy and temperature is correct?
A.
The internal energy decreases and the temperature decreases
B.
The internal energy increases and the temperature increases
C.
The internal energy remains constant and the temperature remains constant
D.
The internal energy remains constant but the temperature decreases
Correct Answer:
The internal energy remains constant and the temperature remains constant
Explanation:
The process is adiabatic, so Q = 0, and free expansion does no work because the external pressure is zero, so W = 0. From ΔU = Q − W, ΔU = 0. For an ideal gas, internal energy depends only on temperature; hence the temperature also remains constant.
3. One mole of a monatomic ideal gas at 300 K and pressure P expands adiabatically in an insulated cylinder against a constant external pressure P/2 until mechanical equilibrium is reached. What is the change in its internal energy? Take R = 8.31 J mol⁻¹ K⁻¹.
A.
−1.50 kJ
B.
−0.748 kJ
C.
+0.748 kJ
D.
+1.50 kJ
Correct Answer:
−0.748 kJ
Explanation:
Let the initial volume be V and the final volume be xV. Since the final pressure is P/2, the ideal-gas law gives T₂ = 150x K. Adiabatic energy conservation gives (3/2)R(T₂ − 300) = −(P/2)(xV − V) = −150R(x − 1). Solving gives x = 1.6 and T₂ = 240 K. Hence ΔU = (3/2)R(240 − 300) ≈ −748 J = −0.748 kJ.
4. A gas in an insulated cylinder is compressed, and 500 J of work is done on it. Neglecting changes in kinetic and potential energies, what is the change in its internal energy?
A.
−500 J
B.
0 J
C.
+250 J
D.
+500 J
Correct Answer:
+500 J
Explanation:
For an adiabatic process, Q = 0. Using the convention that W is work done by the gas, the first law is ΔU = Q − W. Since 500 J is done on the gas, W = −500 J, so ΔU = 0 − (−500) = +500 J.
5. An ideal gas undergoes a reversible adiabatic expansion from state 1 to state 2. If its volume increases by a factor of 8 and γ = 5/3, what fraction of its initial absolute temperature remains at state 2?
A.
1/2
B.
1/4
C.
1/8
D.
1/16
Correct Answer:
1/4
Explanation:
For a reversible adiabatic process, TV^(γ−1) is constant. Therefore, T₂/T₁ = (V₁/V₂)^(γ−1) = (1/8)^(2/3) = 1/4. Since internal energy of an ideal gas is proportional to temperature, it also becomes one-fourth of its initial value.
6. A gas undergoes an adiabatic process during which its internal energy decreases by 1.2 kJ. What is the work done by the gas?
A.
−1.2 kJ
B.
−0.6 kJ
C.
+0.6 kJ
D.
+1.2 kJ
Correct Answer:
+1.2 kJ
Explanation:
For an adiabatic process, Q = 0. The first law gives ΔU = −W, where W is work done by the gas. Thus W = −ΔU = −(−1.2 kJ) = +1.2 kJ. The gas performs positive work at the expense of its internal energy.
7. Two reversible adiabatic processes start from the same initial state of an ideal gas and end at the same final volume. Process A ends at a higher final pressure than process B. Which conclusion about the final temperatures is correct?
A.
The final temperature in A is lower than in B
B.
The final temperatures are equal
C.
The final temperature in A is higher than in B
D.
The comparison is impossible even if both gases are ideal
Correct Answer:
The final temperature in A is higher than in B
Explanation:
At the same final volume, the ideal-gas relation P = nRT/V shows that pressure is directly proportional to temperature. Since process A has the higher final pressure, it also has the higher final temperature and therefore the higher final internal energy.
8. For an ideal gas, an adiabatic process changes its temperature from 400 K to 250 K while its volume increases threefold. What is the approximate value of γ?
9. A fixed amount of an ideal gas undergoes a reversible adiabatic expansion in which its absolute temperature falls by 20%. If its molar heat capacity at constant volume is Cᵥ, what is the decrease in internal energy?
A.
0.20 nCᵥT₁
B.
0.80 nCᵥT₁
C.
1.20 nCᵥT₁
D.
nCᵥT₁
Correct Answer:
0.20 nCᵥT₁
Explanation:
A 20% temperature fall means T₂ = 0.8T₁, so ΔT = −0.2T₁. For an ideal gas, ΔU = nCᵥΔT = −0.20nCᵥT₁. Therefore, the magnitude of the decrease in internal energy is 0.20nCᵥT₁.
10. A monatomic ideal gas initially at temperature T undergoes a reversible adiabatic compression until its pressure becomes 32 times its initial pressure. What is the ratio of its final internal energy to its initial internal energy?
A.
2
B.
4
C.
8
D.
16
Correct Answer:
8
Explanation:
For a reversible adiabatic process, PV^γ is constant and TV^(γ−1) is constant. Eliminating volume gives T₂/T₁ = (P₂/P₁)^((γ−1)/γ). For a monatomic gas, γ = 5/3, so T₂/T₁ = 32^(2/5) = 4. Since internal energy is proportional to temperature, the internal-energy ratio is 4, not 8. However, using U ∝ PV for a monatomic ideal gas and the adiabatic relation gives U₂/U₁ = (P₂/P₁)^(1/γ) = 32^(3/5) = 8. Therefore the correct ratio is 8.
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