Question

A system of ideal gas undergoes isothermal expansion from state (P,V) to a state
(P1, 2V). Find P1 in terms of P. Also find the amount of work done in this process.

Answer 1

In the case of isothermal expansion,

$\sf{\longrightarrow PV=K\quad\quad\dots(1)}$

i.e., it is a constant. Therefore,

$\sf{\longrightarrow PV=P_1(2V)}$

$\sf{\longrightarrow\underline{\underline{P_1=\dfrac{P}{2}}}}$

The work done during this process will be,

$\displaystyle\sf{\longrightarrow W=\int\limits_V^{2V}P\ dV}$

From (1) we can write P as,

$\sf{\longrightarrow P=KV^{-1}}$

Then,

$\displaystyle\sf{\longrightarrow W=\int\limits_V^{2V}KV^{-1}\ dV}$

$\displaystyle\sf{\longrightarrow W=K\Big[\ln V\Big]_V^{2V}}$

$\displaystyle\sf{\longrightarrow W=PV(\ln 2V-\ln V)}$

$\displaystyle\sf{\longrightarrow W=PV\ln\left(\dfrac{2V}{V}\right)}$

$\displaystyle\sf{\longrightarrow\underline{\underline{W=PV\ln2}}}$

Or,

$\displaystyle\sf{\longrightarrow\underline{\underline{W=2.303PV\log2}}}$

$\displaystyle\sf{\longrightarrow W=2.303PV\times0.301}$

$\displaystyle\sf{\longrightarrow\underline{\underline{W=0.693PV}}}$