The pressure of an ideal gas varies with volu | vedclass

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Question

(A) The work done by a gas during expansion is given by the integral $W = \int_{V_1}^{V_2} P dV$.Given the process $P = \alpha V$,where $V_1 = V$ and

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$\frac{\alpha V^2}{2}(m^2-1)$

Vedclass · Answer

(A) The work done by a gas during expansion is given by the integral $W = \int_{V_1}^{V_2} P dV$.Given the process $P = \alpha V$,where $V_1 = V$ and $V_2 = mV$.Substituting the expression for $P$ into the integral:$W = \int_{V}^{mV} (\alpha V') dV'$$W = \alpha \left[ \frac{(V')^2}{2} \right]_{V}^{mV}$$W = \frac{\alpha}{2} [(mV)^2 - V^2]$$W = \frac{\alpha}{2} [m^2 V^2 - V^2]$$W = \frac{\alpha V^2}{2} (m^2 - 1)$

The pressure of an ideal gas varies with volume as $P = \alpha V$,where $\alpha$ is a constant. One mole of the gas is allowed to undergo expansion such that its volume becomes $m$ times its initial volume. The work done by the gas in the process is

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