Question

2L of an ideal gas at a pressure of 10 atm expansion isothermally into a vacuum until its total volume is 10L.

(i) How much heat is absorbed and how much work is done in the expansion?

(ii) How much heat is absorbed if this system expands against a constant external pressure of 1 atm?

(iii) How much heat is absorbed if the expansion is conducted reversibly at 298K.

Answer 1

(i) In case of expansion, work is done by the system.

Now, as we know that,

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

As the gas is expanding is vaccum which has no pressure, i.e.,

$p_{ext.}=0$

$\therefore W=0$

Hence no work is done.

From first law of thermodynamics,

$\Delta U=q+W$

As the system is working at constant temperature, i.e., isothermally.

$\therefore \Delta U=0$

Hence $q=-W=0$

Hence no heat is absorbed in the expansion.

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

$P_{ext}=1atm$

$\Delta V=10-2L=8L$

$\therefore W=-1\times 8=-8atm-L$

As, $q=-W=-(-8)J=8atm-L$

(iii) Now, as the work is done reversibly, then the work done will be given as-

$W_{rev.}=-2.303nRT\log \frac{V_f}{V_i}$

$\Rightarrow W_{rev.}=-2.303PV\log \frac{10}{2}$

$\Rightarrow W_{rev.}=-2.303\times 1\times 10\times \log 5=-16.1L-atm[\because \text{Given}P=1\text{atm in}\left(ii\right)\text{part}]$

As $q_{rev.}=-W_{rev.}$

$q_{rev.}=16.1L-atm$