Chemistry - Thermodynamics - I Question with Solution | TestHub

Explain Question:

Find out correct options

Concept:

Apply general process ${\mathrm{pv}}^{\mathrm{x}}=\mathrm{k}$

Solution:

The process AB is a straight line on he $\mathrm{V}-\mathrm{T}$ graph, so its equation is of the form $\mathrm{V}=\mathrm{mT}+\mathrm{c}$. Using the coordinates of points $\mathrm{A}\left({\mathrm{T}}_{\mathrm{A}}=300\mathrm{K},{\mathrm{V}}_{\mathrm{A}}=10\mathrm{L}\right)$ and $\mathrm{B}\left({\mathrm{T}}_{\mathrm{B}}=600\mathrm{K},{\mathrm{V}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{B}}=15\mathrm{L}\right)$ :

The slope $m$ is

$\mathrm{m}=\frac{{\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{B}}-{\mathrm{T}}_{\mathrm{A}}}=\frac{15\mathrm{L}-10\mathrm{L}}{600\mathrm{K}-300\mathrm{K}}=\frac{5\mathrm{L}}{300\mathrm{K}}=\frac{1}{60}\mathrm{L}/\mathrm{K}$

Using point A to find the intercept c :

${\mathrm{V}}_{\mathrm{A}}={\mathrm{mT}}_{\mathrm{A}}+\mathrm{c}$

$10\mathrm{L}=\left(\frac{1}{60}\mathrm{L}/\mathrm{K}\right)(300\mathrm{K})+\mathrm{c}$

$10\mathrm{L}=5\mathrm{L}+\mathrm{c}$

$\mathrm{c}=5\mathrm{L}$

This matches option (4) $\mathrm{V}=\frac{\mathrm{T}}{60}+5$. We get $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$. We also have $\mathrm{T}=60(\mathrm{V}-5)$ from he line equation, so $\mathrm{dT}=60\mathrm{dV}$

The work done is

$$W_{AB}=-{\int }_{V_A}^{V_B}\frac{nRT}{V}dV$$

$$W_{AB}=-{\int }_{V_A}^{V_B}\frac{nR\cdot 60(V-5)}{V}dV$$

$$W_{AB}=-60nR{\int }_{V_A}^{V_B}\left(1-\frac{5}{V}\right)dV$$

$$W_{AB}=-60nR[V-5\ln |V|{]}_{V_A}^{V_B}$$

Given $n=1mol$, $R=8.314J/mol\cdot K$, $V_A=10L$, $V_B=15L$, $\ln 1.5=0.4$:

$W_{AB}=-1496.52J$

$\Delta H=n{C}_p\Delta T$

$(1mol)\left(\frac{5}{2}\times 8.314J/mol\cdot K\right)(600K-300K)$

Option (3) $\Delta H_{AB}=1982.5J$ is incorrect.

$\Delta U_{AB}=n{C}_V(T_B-T_A)$

$\Delta U_{AB}=(1mol)\left(\frac{3}{2}\times 8.314J/mol\cdot K\right)(600K-300K)$

$\Delta U_{AB}=3741.3J$

$q_{AB}=3741.3J-(-1496.52J)$

$q_{AB}=5237.82J$