1. What is the work done by a gas when it expands against a constant external pressure?
A.
Pressure multiplied by change in volume
B.
Volume divided by pressure
C.
Pressure multiplied by temperature
D.
Change in pressure multiplied by volume
Correct Answer:
Pressure multiplied by change in volume
Explanation:
For expansion against constant pressure, the work done by the gas is W = PΔV, where P is the pressure and ΔV is the change in volume.
2. A gas expands from 2 m³ to 5 m³ against a constant pressure of 100 Pa. What work is done by the gas?
A.
200 J
B.
300 J
C.
500 J
D.
700 J
Correct Answer:
300 J
Explanation:
The change in volume is 5 − 2 = 3 m³. Therefore, W = PΔV = 100 × 3 = 300 J.
3. When a gas is compressed, the work done by the gas is generally
A.
Positive
B.
Negative
C.
Always zero
D.
Always equal to its internal energy
Correct Answer:
Negative
Explanation:
During compression, the volume decreases, so ΔV is negative. Using W = PΔV, the work done by the gas is negative.
4. What is the work done by a gas if its volume remains constant?
A.
Zero
B.
Equal to pressure
C.
Equal to temperature
D.
Maximum
Correct Answer:
Zero
Explanation:
At constant volume, the change in volume is zero. Hence, W = PΔV = 0.
5. A gas expands by 4 m³ at a constant pressure of 50 Pa. What is the work done by the gas?
A.
12.5 J
B.
54 J
C.
200 J
D.
250 J
Correct Answer:
200 J
Explanation:
Using W = PΔV, the work done is 50 × 4 = 200 J.
6. In a pressure-volume graph, the work done by a gas during expansion is represented by the area
A.
Below the curve and above the volume axis
B.
Above the pressure axis only
C.
Inside the temperature scale
D.
Outside the graph completely
Correct Answer:
Below the curve and above the volume axis
Explanation:
The work done is W = ∫P dV, which is represented by the area under the pressure-volume curve.
7. A gas expands from 1 L to 3 L against a constant pressure of 2 × 10⁵ Pa. What is the work done by the gas?
A.
200 J
B.
400 J
C.
600 J
D.
800 J
Correct Answer:
400 J
Explanation:
The volume change is 2 L = 2 × 10⁻³ m³. Thus, W = PΔV = 2 × 10⁵ × 2 × 10⁻³ = 400 J.
8. Which quantity is directly required, along with pressure, to calculate work done during a constant-pressure expansion?
A.
Change in volume
B.
Change in mass
C.
Change in density only
D.
Change in temperature only
Correct Answer:
Change in volume
Explanation:
For constant pressure, the work done is given by W = PΔV, so the change in volume is required.
9. If a gas expands against zero external pressure, the work done by the gas against the external pressure is
A.
Zero
B.
Equal to its pressure
C.
Equal to its volume
D.
Always positive and maximum
Correct Answer:
Zero
Explanation:
Work against an external pressure is W = PextΔV. If the external pressure is zero, the work done against it is zero.
10. A gas is compressed from 6 m³ to 4 m³ at a constant pressure of 100 Pa. What is the work done by the gas?
A.
-200 J
B.
-100 J
C.
200 J
D.
400 J
Correct Answer:
-200 J
Explanation:
The volume change is 4 − 6 = −2 m³. Therefore, W = PΔV = 100 × (−2) = −200 J.
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