Question

Calculate the maximum work done when pressure on $10g$ of hydrogen is reduced from $20$ to $1atm$ at a constant temperature of $273K$. The gas behaves ideally. Calculate $Q$.

Answer 1

Solution:-

We can get maximum work done when the process is reversible.

Given:-

$P_1=20atm$

$P_2=1atm$

$T=273K$

Weight of $H_2=2g$

$\Rightarrow$ No. of moles of $H_2=\frac{10}{2}=5\text{mol}$

Now as we know that, the work done in an isothermal reversible process-

$W=-2.303nRT\log \frac{P_i}{P_f}$

$\therefore W=-2.303\times 5\times 2\times 273\times \log \frac{20}{1}$

$\Rightarrow W=-8.18kcal$

For an ideal gas, in isothermal condition, change in internal energy is zero, i.e.,

$\Delta U=0$

From first law of thermodynamics,

$\Delta U=q+W$

$\because \Delta U=0$

$\therefore q=-W=-(-8.18)=8.18kcal$