Question

Calculate the $Q$ for the isothermal reversible expansion of 1 mole of an ideal gas from initial pressure of 1.0 bar to a final pressure 0.1 bar at a constant temperature at 273 K.

Answer 1

Given :- $P_1=1bar$ $P_2=0.1bar$ $T_c=273K$

$n=1mole$; Isothermal process $T=const$

$\Delta U=Q+W$ ---- in isothermal process $\Delta U=0$

$\Rightarrow Q=-W$

Now,

$W=-P_1{V}_{1}ln\frac{V_2}{V_1}$ or $-nRTln$ $\frac{V_2}{V_1}\Rightarrow W=$ $-nRT\frac{P_1}{P_2}$

$W=-1\times 8.314\times 273\times ln\left(\frac{1}{0.1}\right)$

$W=-8.314\times 273\times ln10$

$W=-5226.23J$

$\therefore Q=5226.23J$