The pressure of an ideal gas varies with volume as $P=\alpha V$, where $\alpha$ is a constant. One mole of…

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
  1. $\frac{\alpha V}{2}\left(m^2-1\right)$
  2. $\frac{\alpha^2 V^2}{2}\left(m^2-1\right)$
  3. $\frac{\alpha}{2}\left(m^2-1\right)$
  4. $\frac{\alpha V^2}{2}\left(m^2-1\right)$

Solution

Given $P=\alpha V$ Work done, $w=\int_V^{m V} P d V$ $=\int_V^{m V} \alpha V d V=\frac{\alpha V^2}{2}\left(m^2-1\right)$.

Asked in: JEE Main 2012 (19 May Online)

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