2. A monatomic ideal gas occupying a rigid vessel has an initial pressure of 1.2 atm at 300 K. It is heated until its pressure becomes 2.0 atm. If the gas amount is 0.5 mol, its increase in internal energy is:
A.
75R
B.
100R
C.
150R
D.
250R
Correct Answer:
150R
Explanation:
At constant volume, P/T is constant, so the final temperature is T₂ = 300 × (2.0/1.2) = 500 K. Therefore, ΔU = 0.5 × (3R/2) × 200 = 150R.
3. One mole of a diatomic ideal gas, with vibrational modes neglected, is heated in a rigid vessel from 300 K to 500 K. The heat supplied is approximately:
A.
2.08 kJ
B.
3.12 kJ
C.
4.16 kJ
D.
6.24 kJ
Correct Answer:
4.16 kJ
Explanation:
For a diatomic ideal gas, Cᵥ = 5R/2. Since the volume is constant, W = 0 and Q = ΔU = nCᵥΔT = (5R/2)(200) ≈ 4.16 kJ.
4. A rigid container holds 1 mol of a monatomic ideal gas and 2 mol of a diatomic ideal gas. Both gases are heated through the same temperature increase of 100 K, with vibrational modes neglected. The total increase in internal energy is:
A.
400R
B.
500R
C.
650R
D.
800R
Correct Answer:
650R
Explanation:
The total heat capacity at constant volume is 1(3R/2) + 2(5R/2) = 13R/2. Hence, ΔU = (13R/2)(100) = 650R.
5. A fixed amount of monatomic ideal gas is heated at constant volume until its pressure becomes three times its initial value. If its initial temperature is 280 K, the ratio of the final internal energy to the initial internal energy is:
A.
1:3
B.
3:1
C.
1:9
D.
9:1
Correct Answer:
3:1
Explanation:
For a fixed amount of gas at constant volume, P is proportional to T. Thus, the final temperature is three times the initial temperature. Since internal energy of an ideal gas is proportional to temperature, U₂/U₁ = 3.
6. A monatomic ideal gas is heated in a rigid container from pressure 2.0 × 10⁵ Pa to 5.0 × 10⁵ Pa. The volume of the container is 3.0 × 10⁻³ m³. What is the increase in internal energy?
A.
450 J
B.
900 J
C.
1350 J
D.
2250 J
Correct Answer:
1350 J
Explanation:
For a monatomic ideal gas, U = 3PV/2. At constant volume, ΔU = (3/2)VΔP = (3/2)(3.0 × 10⁻³)(3.0 × 10⁵) = 1350 J.
7. A monatomic ideal gas in a rigid vessel initially has pressure 4.0 × 10⁵ Pa and volume 1.0 × 10⁻² m³. It is cooled until its pressure decreases by 20%. The change in internal energy is:
8. Equal amounts of heat are supplied separately to equal numbers of moles of a monatomic ideal gas and a diatomic ideal gas in rigid containers. If the vibrational modes of the diatomic gas are neglected, the ratio of their temperature rises, ΔT_monatomic : ΔT_diatomic, is:
A.
2:3
B.
3:5
C.
5:3
D.
1:1
Correct Answer:
5:3
Explanation:
At constant volume, Q = nCᵥΔT. For equal n and Q, temperature rise is inversely proportional to Cᵥ. Since Cᵥ(monatomic) = 3R/2 and Cᵥ(diatomic) = 5R/2, the ratio is ΔT_mono/ΔT_diatomic = 5/3.
9. A gas is heated in a perfectly rigid, thermally insulated-from-work container. During the process, 2.5 kJ of energy is transferred to the gas as heat. Which statement correctly describes the process?
A.
The gas does 2.5 kJ of work and its internal energy remains unchanged
B.
The internal energy increases by 2.5 kJ because the work done is zero
C.
The internal energy decreases by 2.5 kJ because the volume is constant
D.
The pressure remains constant and the internal energy increases by 1.25 kJ
Correct Answer:
The internal energy increases by 2.5 kJ because the work done is zero
Explanation:
For an isochoric process, ΔV = 0, so the work done by the gas W = PΔV = 0. The first law therefore gives ΔU = Q = 2.5 kJ.
10. One mole of a monatomic ideal gas undergoes a constant-volume process in which its temperature changes from T to 4T. If its initial internal energy is U, the heat supplied during the process is:
A.
U
B.
2U
C.
3U
D.
4U
Correct Answer:
3U
Explanation:
For a monatomic ideal gas, internal energy is proportional to temperature. Increasing the temperature from T to 4T increases the internal energy from U to 4U, so ΔU = 4U − U = 3U. At constant volume, Q = ΔU.
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