Question

A sample of ideal gas is expanded to twice its original volume of $1.00m^3$ in a quasi-static process for which $P=\alpha V^2$, with $\alpha =5.00atm/m^6$, as shown in figure. How much work is done by the expanding Gas ?

• A. $11.33\times {10}^{5}J$
• B. $18.33\times {10}^{5}J$
• C. $10.33\times {10}^{5}J$
• D. $3.33\times {10}^{5}J$

Answer 1

The correct option is A $11.33\times {10}^{5}J$

Given : $V_1=1m^3$, $V_2=2m^3$ and $\alpha =5$ $\frac{atm}{m^6}=5\times {10}^5$ $\frac{Pa}{m^6}$

The process is given as $P=\alpha V^2$

$\therefore$ Work done by the gas $W=\int PdV$

$⟹$ $W={\int }_{V_1}^{V_2}\alpha V^{2}dV$

$⟹$ $W=\frac{\alpha }{3}\times V^3{|}_{V_1}^{V_2}$

$⟹$ $W=\frac{\alpha }{3}\times (V_2^3-V_1^3)$

$⟹$ $W=\frac{5\times {10}^5}{3}\times (2^3-1^3)=11.33\times {10}^5$ J