Question

Two liters of an ideal gas at a pressure of 10 atm expands isothermally into vacuum until its total volume is 10 liters. How much heat is absorbed and how much work is done in the expansion? What would be the heat absorbed and work done?
(i) if the same expansion takes place against a constant external pressure of 1 atm?
(ii) if the same expansion takes place to a final volume of 10 liters conducted reversibly?

Answer 1

We know that in the expansion of any Gas, work is done by the system

$W=-P_{ext}\Delta v=-P_{ext}\left(V_f-V_i\right)$

$p_{ext}$ $=$ external pressure of the system $=0$

In this Problem Pressure of $10atm$ is given and it is the pressure of Gas itself but we required external pressure in the expression and the gas is expanding into vacuum which has no pressure hence the external pressure $=0$

$\begin{array}{c}{V}_f=10L\\ V_i=2L\end{array}$

$\because workdonebythesystem$

$\begin{array}{c}W=-0\times \left(10-2\right)=0\end{array}$

Hence, No work is done and heat absorbed $\left(q\right)$ is a consequence of gaining temperature, but the system is working at constant temperature (Isothermal Condition). $q=-w$ so, there will be no heat absorbed

$\left(i\right)$

$\begin{array}{c}{P}_{ext}=1atm\\ V_f=10L\\ V_i=2L\\ \because W=-1atm\times \left(10-2\right)L=-8atmL\end{array}$

Also, during isothermal expansion $q=-w=8atmL$

$\left(ii\right)$

Now the work is done then the expression of work done is

$W_{rev}={\int }_{V_i}^{V_f}{P}_{in}dv$

$P_{in}$ is the pressure of the Gas and not external pressure

Since, the gas is ideal, so $PV=nRT$ or $\begin{array}{c}{P}_{in}=\frac{nRT}{V}\end{array}$

$\begin{array}{c}W={\int }_{V_i}^{V_f}nRT\frac{dV}{V}\\ =-nRTIn\frac{V_f}{V_i}\\ =-2.303nRT\log \frac{V_f}{V_i}\end{array}$

so,

$P_{in}=10atm=\frac{nRT}{V}$

$\begin{array}{c}{V}_f=10L\\ V_i=2L\end{array}$

$\begin{array}{c}W=-2.303nRT\log \frac{V_f}{V_i}\\ =-2.303\times 20\times \log \frac{10}{2}\\ =-32.2atmL\\ Q=-W=32.2atm.L\end{array}$