# One mole of an ideal gas at temperature T1 expands slowly according to the law p/v =constant.The final temperature is T2 the workdone by the gas is

Grade:12

## 2 Answers

Khimraj
3007 Points
4 years ago
Let P/V = a(constant)
then work done = $\int Pdv$ = $\intaVdv$$\intaVdv$$\int_{v1}^{v2}avdv$ = (a/2)(v22 – v12)
Now PV = nRT
n = 1,
P = aV
So V2 = RT/a
So v22 = RT2/a
and V12 = RT1/a
So work done
W = R(T2 – T1)/2
Rishi Sharma
askIITians Faculty 646 Points
2 years ago
Dear Student,
Please find below the solution to your problem.

One mole of an ideal gas at temperature t1 expands according to the law p/v =constant.The final temperature is T2.
Suppose P/V = a(constant)
then work done = \int Pdv = \int aVdv\int aVdv\int_{v1}^{v2}avdv = (a/2)(v2square – v1square)
Now PV = nRT , n = 1,
PV = RT
P = aV
V square = RT/a
So v2 square = RT2/a
and V1 square = RT1/a
work done: W = R(T2 – T1)/2

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