Question

Calculate the entropy change accompanying the following change of state

$5mol$ of $O_2({27}^{\circ }C,1atm)\to 5mol$ of $O_2({117}^{\circ }C,5atm)$

$C_P$ for $O_2=6.95calde{g}^{-1}mo{l}^{-1}$

• A. $-8.88calde{g}^{-1}$
• B. $-6.88calde{g}^{-1}$
• C. $-9.77calde{g}^{-1}$
• D. $-10.77calde{g}^{-1}$

Answer 1

The correct option is B $-6.88calde{g}^{-1}$

Here lets devide the process in 2 stages

$\Delta S_1\to 5mol$ of $O_2\left(1atm\right)\to 5mol$ of $O_2\left(5atm\right)$

$\Delta S_2\to T\to {27}^{\circ }$ to ${117}^{\circ }C$

Now,

$\Delta S_1=\frac{q_{rev}}{T}=\frac{-W_{rev}}{T}=\frac{1}{T}\times {\int }_{V_1}^{V_2}PdV$

= $\frac{1}{T}.nRT{\int }_{V_1}^{V_2}\frac{dV}{V}=nR\times 2.303log\frac{V_2}{V_1}$

= $5\times 1.987\times 2.303log\frac{P_1}{P_2}$

= $5\times 1.987\Delta 2.303log\frac{1}{5}=-16.0calde{g}^{-1}$

$\Delta S_2$ = for the change of temperature from ${27}^{\circ }C$ to ${117}^{\circ }C$

$\Delta S_2=n{C}_P{\int }_{T_1}^{T_2}\frac{dT}{T}$

= $5\times 6.95\times 2.303log\frac{390}{300}$

= $5\times 6.95\times 2.303\times 0.1139$

= $9.12calde{g}^{-1}$

$\Delta S=-16.0+9.12=-6.88calde{g}^{-1}$