Question

Four moles of an ideal gas at $2.5$ atm and ${27}^0$C are compressed isothermally to half of its volume by an external pressure of $3$ atm. Calculate w, q and $△U$.

Answer 1

Given,

Pressure of ideal gas $\left(P\right)=2.5atm$

Temperature of ideal gas $\left(T\right)=27℃$

No. of moles $\left(n\right)=4\text{moles}$

Initial volume of gas $\left(\right)=V=?$

Now, from ideal gas equation,

$PV=nRT$

$2.5\times V=4\times 0.0821\times 300$

$\Rightarrow V=39.408L$

Final volume $\left(V_2\right)=\frac{V}{2}=\frac{39.408}{2}=19.704L$

$\Delta V=V_2-V_1=19.704-39.408=-19.704L$

$P_{ext.}=3atm\left(\text{Given}\right)$

Now, as we know that,

$W=-P_{ext.}\Delta V$

$\therefore W=-3\times (-19.704)=59.112L-atm$

As thw process is isothermal,

$\Rightarrow \Delta U=0$

Now from first law of thermodynamics, i.e.,

$\Delta U=q+W$

$\because \Delta U=0$

$\therefore q=-W=-59.112L-atm$.