Question

Calculate the final volume of one mole of an ideal gas initially at $0$ and 1 atm pressure, if it absorbs 1000 cal of heat during a reversible isothermal expansion.

Answer 1

Solution:-

The gas is in the standard temperature and pressure condition i.e. at S.T.P

Therefore,

$V_i=22.4{dm}^3$

$V_f=?$

As given that the expansion is isothermal and reversible

$\therefore \Delta U=0$

Now from first law of thermodynamics,

$\Delta U=q+w$

$\because \Delta U=0$

$\therefore q=–w$

Given that the heat is absorbed.

$\therefore q=1000cal$

$\Rightarrow w=-q=-1000cal$

Now,

Work done in a reversible isothermal expansion is given by-

$w=-nRT\ln \left(\frac{V_f}{V_i}\right)$

Given:-

$T=0℃=273K$

$n=1\text{mol}$

$\therefore 1000=-nRT\ln \left(\frac{V_f}{V_i}\right)$

$\Rightarrow 1000=-1\times 2.303\times 2\times 273\times \log \left(\frac{V_f}{22.4}\right)$

$\Rightarrow \log \left(\frac{V_f}{22.4}\right)=-\frac{1000}{2.303\times 273\times 2}$

$\Rightarrow \log \left(\frac{V_f}{22.4}\right)=-0.8$

$\Rightarrow V_f=3.5{dm}^3$

Hence the final volume will be $3.5{dm}^3$.