Question

A sample of an ideal gas is expanded $1m^3$ to $3m^3$ in a reversible process for which $P=K{V}^2$, with $K=6bar/m^6$. Work done by the gas is:

• A. $5200kJ$
• B. $15600kJ$
• C. $52kJ$
• D. $5267.6kJ$

Answer 1

The correct option is A $5200kJ$

Solution:- (A) $5200kJ$

$W=-{\int }_{V_i}^{V_f}PdV$

Given:-

$V_i=1m^3$

$V_f=3m^3$

$P=K{V}^2=(6\times {10}^5)V^{2}Pa/m^6(K=6bar/m^6=6\times {10}^{5}Pa/m^6)$

$\therefore W=-{\int }_1^3(6\times {10}^5)V^{2}dV$

$\Rightarrow W=-6\times {10}^5{\left[\frac{V^3}{3}\right]}_1^3$

$\Rightarrow W=-\frac{6\times {10}^5}{3}(3^3-1^3)$

$\Rightarrow W=-\frac{6\times {10}^5}{3}\times 26$

$\Rightarrow W=-52\times {10}^{5}J=-5200kJ$

Here $-ve$ sign indicates that the work is done by the gas.

Hence the work done by the gas is $5200kJ$