Question

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

Answer 1

Given $P=\alpha V^2$
Where $\alpha =5{\text{atm/m}}^6$
Initial volume, $V_i=1{\text{m}}^3$
Final volume, $V_f=2V_i=2{\text{m}}^3$
Work done by the gas
$W={\int }_i^{f}PdV={\int }_i^f\alpha V^{2}dV=\frac{1}{3}\alpha (V_f^3-V_i^3)$
$W=\frac{1}{3}\left[\right(5.00{\text{atm/m}}^6\left)\right(1.013\times {10}^5\text{Pa/atm}\left)\right]\times \left[\right(2.00{\text{m}}^3{)}^3-\left(1.00{\text{m}}^3{)}^3\right]$
$=1.18\text{MJ}$