Question

For the reversible isothermal expansion of one mole of an ideal gas at 300 K, from the volume of $10{dm}^3$ to $20{dm}^3,\Delta H$ is:

• A. $5.73kJ$
• B. $1.73kJ$
• C. $3.46kJ$
• D. Zero

Answer 1

Answer: (B)

$1.73kJ$

Solution:-

Work done in an isothermal reversible process is given by-

$W=-nRT\ln \frac{V_2}{V_1}$

Where,

$n$ = no. of moles $=1$

$R=$ molar gas constant $=8.314J{mol}^{-1}{K}^{-1}$

$T=$ temperature $=300K$

$V_1=10{dm}^3$

$V_2=20{dm}^3$

$\therefore W=-1\times 8.314\times 300\times \ln \frac{20}{10}$

$\Rightarrow W=-2494.2\times (\ln 2-\ln 1)=-1728.8J=-1.73kJ$

Now, from first law of thermodynamics,

$\Delta U=\Delta H+W$

Since temperature is constant, i.e. $\Delta U=0$.

$\therefore \Delta H=-W=-(-1.73)=+1.73KJ$