Question

In a process the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gas

• A. is positive
• B. is negative
• C. is zero
• D. may be positive or negative

Answer 1

The correct option is A is positive

Given that
$P\propto V^2$
$P=k{V}^2$
$PV=nRT$
We get $V=k_2{T}^3$
If the temperature increases, so does the volume, Work done, $w=\int PdV=\int k{V}^{2}dV=k\frac{V^3}{3}+c$
Between the limits $V_1$ to $V_2$
$w=k(\frac{V_2^3}{3}-\frac{V_1^3}{3})>0$
Hence work done is positive.
Hence option A is correct.