The pressure and volume of an ideal gas are r | vedclass

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Question

(A) For a polytropic process $PV^x = K$,the work done by the gas is given by the formula:$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} K V^{-x} \,

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$2(P_1 V_1 - P_2 V_2)$

Vedclass · Answer

(A) For a polytropic process $PV^x = K$,the work done by the gas is given by the formula:$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} K V^{-x} \, dV$$W = \frac{K V_2^{1-x} - K V_1^{1-x}}{1-x}$Since $P_1 V_1^x = K$ and $P_2 V_2^x = K$,we can write:$W = \frac{P_2 V_2^x V_2^{1-x} - P_1 V_1^x V_1^{1-x}}{1-x} = \frac{P_2 V_2 - P_1 V_1}{1-x}$Given $x = 3/2$,we substitute this into the equation:$W = \frac{P_2 V_2 - P_1 V_1}{1 - 3/2} = \frac{P_2 V_2 - P_1 V_1}{-1/2}$$W = -2(P_2 V_2 - P_1 V_1) = 2(P_1 V_1 - P_2 V_2)$Thus,the correct option is $A$.

The pressure and volume of an ideal gas are related as $PV^{3/2} = K$ (constant). The work done when the gas is taken from state $A(P_1, V_1, T_1)$ to state $B(P_2, V_2, T_2)$ is:

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