2004 AIPMT ThermodynamicsWork in Irreversible Expansion Easy Question 67 of 82 in Thermodynamics

The work done during the expansion of a gas from a volume of 4 dm$^3$ to 6 dm$^3$ against a constant external pressure of 3 atm is :-

A –608 J
B +304 J
C –304 J
D –6 J

The answer is A.

Solution

For expansion against a constant external pressure: $W = -P_{ext}\Delta V$ (work done by the gas is negative in the IUPAC sign convention).
$\Delta V = 6 - 4 = 2$ dm$^3$ $= 2$ L
$W = -3$ atm $\times 2$ L $= -6$ L atm
1 L atm $= 101.3$ J
$W = -6 \times 101.3 = -607.8$ J $\approx -608$ J
So the work done is $-608$ J.

Concept Pressure-volume work

Thermodynamics in past papers

82 questions from this chapter have appeared across 26 exam years.