An ideal gas is taken through a quasi-static | vedclass

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Question

(B) The work done by the gas in a quasi-static process is given by $W = \int_{V_1}^{V_2} P \, dV$.Given $P = \alpha V^2$ and $\alpha = 5 	ext{ atm}/m

Vedclass · Accepted Answer

$1.18 	ext{ MJ}$

Vedclass · Answer

(B) The work done by the gas in a quasi-static process is given by $W = \int_{V_1}^{V_2} P \, dV$.Given $P = \alpha V^2$ and $\alpha = 5 	ext{ atm}/m^6$. Since $1 	ext{ atm} = 1.01325 	imes 10^5 	ext{ Pa}$,we use $\alpha = 5 	imes 1.01325 	imes 10^5 	ext{ Pa}/m^6 \approx 5.066 	imes 10^5 	ext{ Pa}/m^6$.Initial volume $V_1 = 1 	ext{ m}^3$,final volume $V_2 = 2 	ext{ m}^3$.$W = \int_{1}^{2} (\alpha V^2) \, dV = \alpha \left[ \frac{V^3}{3} ight]_1^2 = \frac{\alpha}{3} (2^3 - 1^3) = \frac{7\alpha}{3}$.Substituting $\alpha = 5.066 	imes 10^5 	ext{ Pa}/m^6$:$W = \frac{7 	imes 5.066 	imes 10^5}{3} \approx 11.82 	imes 10^5 	ext{ J} = 1.182 	ext{ MJ}$.Thus,the correct option is $B$.

An ideal gas is taken through a quasi-static process described by $P = \alpha V^2$,where $\alpha = 5 \text{ atm}/m^6$. The gas is expanded to twice its original volume of $1 \text{ m}^3$. How much work is done by the expanding gas in this process?

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