Question

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

• A. $6362.8$J, $4700$J
• B. $1522.2$ Cal, $1124.4$ Cal
• C. $7062.8$, $5400$ J
• D. $3181.4$ J, $2350$ J

Answer 1

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

Solution:- (A) $6362.8J,4700J$

$dU=n{C}_{v,m}dT$

Given:-

$C_{v,m}=20+{10}^{-2}TJ/K-mol$

$\therefore dU=n(20+{10}^{-2}T)dT$

Integrating the above equation, we have

${\int }_{U_1}^{U_2}dU={\int }_{T_1}^{T_2}n(20+{10}^{-2}T)dT$

Given:-

$T_1=27℃=(27+273)=300K$

$T_2=127℃=(127+273)=400K$

$\therefore (U_2-U_1)=n{[20T+{10}^{-2}\frac{T^2}{2}]}_{300K}^{400K}$

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

$\Rightarrow \Delta U=4700J$

Now as we know that, when pressure is constant,

$\Delta U=q-P\Delta V$

From ideal gas equation-

$PV=nRT$

$\Rightarrow \Delta V=\frac{nR}{P}\Delta T$

$\therefore \Delta U=q-nR\Delta T$

$\Rightarrow q=\Delta U+nR\left(\Delta T\right)$

$\Rightarrow q=4700+(2\times 8.314\times 100)=6362.8J$

Hence value of $q$ and $\Delta U$ are $6362.8J$ and $4700J$ respectively.