Question

An ideal gas is taken through a quasi-static process described by $P=\alpha V^{2}$, with $\alpha=5.00\, atm / m ^{6} .$ The gas is expanded to twice its original volume of $1.00\, m ^{3}$. How much work is done by the gas in expanding gas in this process?

• A. 1.18 MJ
• B. 2.18 MJ
• C. 1.28 MJ
• D. 3.18 MJ

Answer 1

Answer: (A)

The work done by the gas
$W =\int\limits_{i}^{f} P d V$
$W =\int\limits_{i}^{f} \alpha V^{2} d V=\frac{1}{3} \alpha\left(V_{f}^{3}-V_{i}^{3}\right)$
$V_{f} =2 V_{i}=2\left(1.00\, m ^{3}\right)=2.00 \,m ^{3}$
$W =\frac{1}{3} \left[\left(5.00\, atm / m ^{6}\right)\left(1.013 \times 10^{5} Pa / atm \right)\right]$
$\times\left[\left(2.00 \,m ^{3}\right)^{3}-\left(1.00\, m ^{3}\right)^{3}\right]=1.18\, MJ$