In a cyclical process on a pressure-volume diagram, a gas moves along segment AB at a constant pressure of 1.50 x 10^6 N/m^2 while its volume increases by 5.00 x 10^-4 m^3. What is the work done by the gas along this segment?
A100 J
B750 J
C850 J
D1500 J
Answer & Solution
Correct answer: B. 750 J
1. Segment AB is an isobaric process, so the work done by the gas is W = P x ΔV.
2. Substitute the given values with their units: W_AB = (1.50 x 10^6 N/m^2)(5.00 x 10^-4 m^3).
3. Multiplying the numbers: 1.50 x 5.00 = 7.50, and the powers of ten combine as 10^6 x 10^-4 = 10^2.
4. So W_AB = 7.50 x 10^2 N*m = 750 J, since one newton-meter equals one joule.
5. Option A, 100 J, would come from mistakenly using only the power-of-ten exponent difference without multiplying the leading numbers.
6. Option C, 850 J, and option D, 1500 J, do not follow from correctly multiplying 1.50 x 10^6 by 5.00 x 10^-4, so neither matches the required unit-by-unit substitution.
_Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.2 The First Law of Thermodynamics and Some Simple Processes_
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