10.
In a certain process 1.5×105 Joules of heat is added to an ideal gas to keep the pressure at
2.0×105 Pa while the volume expands from 6.3m3 to 7.1m3. What is the change in
internal energy for the gas?
a. It decreases by 2.0×105 J
b. It increases by 1.0×105 J
c. It increases by 1.0×103 J
d. It decreases by 1.0×104J
Introduction
This article explains how to calculate the change in internal energy of an ideal gas during isobaric expansion using the first law of thermodynamics, applied to a numerical multiple-choice question. The step-by-step solution clarifies the roles of heat, work, and internal energy, followed by a detailed analysis of each option.
Given Data and Concept
- Heat added to the gas: Q = 1.5 × 105 J
- Constant pressure: P = 2.0 × 105 Pa
- Volume change: from V1 = 6.3 m3 to V2 = 7.1 m3
- Process: isobaric (constant pressure)
First law of thermodynamics (physics sign convention):
ΔU = Q – W
where W is work done by the gas.
For an isobaric process, work done by the gas is:
W = P ΔV = P (V2 – V1)
Step-by-Step Calculation
Calculate ΔV
ΔV = V2 – V1 = 7.1 – 6.3 = 0.8 m3
Calculate work done by the gas
W = P ΔV = (2.0 × 105 Pa)(0.8 m3) = 1.6 × 105 J
Apply the first law
ΔU = Q – W = 1.5 × 105 – 1.6 × 105 = -0.1 × 105 J = -1.0 × 104 J
Conclusion: The internal energy decreases by 1.0 × 104 J.
The change in internal energy of the gas is a decrease of 1.0 × 104 J, so the correct option is (d) It decreases by 1.0 × 104 J.
Explanation of Each Option
(a) It decreases by 2.0 × 105 J
This would imply |ΔU| larger than both the heat added and the work done, which contradicts ΔU = Q – W = -1.0 × 104 J. Therefore, option (a) is incorrect.
(b) It increases by 1.0 × 105 J
An increase would require Q > W by 1.0 × 105 J, but here W > Q and the gas does more work than the heat it absorbs, so ΔU is negative, not positive. Hence, option (b) is incorrect.
(c) It increases by 1.0 × 103 J
This again assumes a positive ΔU of small magnitude, which is inconsistent with the calculated negative value -1.0 × 104 J. Thus, option (c) is incorrect.
(d) It decreases by 1.0 × 104 J
This matches the computed result ΔU = -1.0 × 104 J, meaning the gas loses internal energy because work done exceeds heat absorbed. Therefore, option (d) is correct.
