Question

13. One mole of an ideal gas is heated reversibly along a path such that T= AV?, where A is a constant. If the initial temperature is T(K), what must the final temperature T (K)) be if the entropy change is equal to BOK-) (B is a constant value and cm = R (Jmol-'K™)

Answer 1

For, entropy change of an ideal gas the equation is,

$\Delta S= nC_v^mln \frac{T_f}{T_i}+nR ln \frac{V_f}{V_i}$

where, i stands for initial and f stands for final.

Now, T_i = T₁, $\Delta S$ = B and

$V^2=\frac{T}{A}\Rightarrow V=(\frac{T}{A})^{1/2}$

$Thus,\frac{V_f}{V_i}=(\frac{T_f}{T_1})^{1/2}$

So, we get from the equation,

$B (J.K^{-1}) = 1mol\times \frac{3}{2} R (J.mol^{-1}.K^{-1})\times ln\frac{T_f}{T_1}+1 mol\times R(J.mol^{-1}.K^{-1})\times ln(\frac {T_f}{T_1})^{1/2}$

$B= \frac{3}{2} R \times ln\frac{T_f}{T_1}+ \frac{R}{2}\times ln(\frac {T_f}{T_1})$

$Thus,B= (\frac{3R}{2} + \frac{R}{2})\, ln(\frac {T_f}{T_1})$

$Thus,\frac{B}{2R}=ln(\frac {T_f}{T_1})$

$Thus,\frac {T_f}{T_1}=e^{\frac{B}{2R}}$

$Thus,T_f=T_1e^{\frac{B}{2R}}$

So, final temperature should be this much Kelvin.