Question

Two mole of an ideal gas is heated at constant pressure of one atmosphere from ${27}^{\circ }C$ to ${127}^{\circ }C$. If $C_{v.m}=20+{10}^{-2}TJ{K}^{-1}mo{l}^{-1}$, then $q$ and $△U$ for the process are respectively.

• A. $6362.8J,4700J$
• B. $3037.2J,4700J$
• C. $7062.8J,5400J$
• D. $3181.4J,2350J$

Answer 1

The correct option is A $6362.8J,4700J$

Given,

Moles of ideal gas $=2$

Constant pressure process

Initial temperature $\left(T_1\right)={27}^{o}C=300K$

Final temperature $\left(T_2\right)={127}^{o}C=400K$

$C_v=20+{10}^{-2}T\frac{J}{Kmol}$

$C_p=R+20+{10}^{-2}T\frac{J}{molK}$

To find, $q$ and $\Delta U$ for the process.

$\Delta U=n{C}_v\Delta T$

$dU=n{C}_{v}dt$ $[\therefore$ on intergrating$]$

${\int }_{v_1}^{v_2}dU=n{\int }_{T_1}^{T_2}{C}_{v}dT$

$U_2-U_1=n\left[20\right(T_2-T_1)+\frac{{10}^{-2}}{2}(T_2^2-T_1^2\left)\right]$

$\Delta U=2\times \left[20\right(400-300)+\frac{{10}^{-2}}{2}({400}^2-{300}^2\left)\right]$

$\Delta U=4700J$

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

$dq=n{C}_{p}dT$ [ on agian integrating]

${\int }_{q_1}^{q_2}dq=n{\int }_{T_1}^{T_1}{C}_{p}dT$

$q_2-q_1=n\left[R\right(T_2-T_1)+20(T_2-T_1)+\frac{{10}^{-2}}{2}(T_2^2-T_1^2\left)\right]$

$\Delta q=2\times \left[8.314\right(400-300)+20(400-300)+\frac{{10}^{-2}}{2}({400}^2-{300}^2\left)\right]$

$\Delta q=6362.8J$

$\therefore$ option A is correct