Question

$1$ mole of an ideal gas $A(C_{v,m}=3R)$ and $2$ mole of an ideal gas $B$ are $\left(C_{v,m}\frac{3}{2}R\right)$ taken in a container and expanded reversible and adiabatically from $1$ litre to $4$ litre starting from initial temperature of $320$ K. $△Eor△U$ for the process is:

• A. $-240R$
• B. $240R$
• C. $480R$
• D. $-960R$

Answer 1

The correct option is D $-960R$

Solution:- (D) $-960R$

As we know that,

$C_v=\frac{f}{2}R$

$\Rightarrow f=\frac{2C_v}{R}$

Given:- ${C_{v,m}}_A=3\Rightarrow f_A=6$

${C_{v,m}}_B=\frac{3}{2}R\Rightarrow f_B=3$

$n_A=1\text{mole}$

$n_B=2\text{mole}$

By using-

$f_{av.}=\frac{f_A{n}_A+f_B{n}_B}{n_A+n_B}$

$\Rightarrow f_{av.}=\frac{1\times 6+2\times 3}{1+2}$

$\Rightarrow f_{av.}=4$

${\gamma }_{av.}=1+\frac{2}{f_{av.}}$

$\Rightarrow {\gamma }_{av.}=1+\frac{2}{4}=\frac{3}{2}$

Now as we know that, for an adiabatic process,

$T{V}^{{\gamma }_{av.}-1}=\text{constant}$

$\Rightarrow \frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\frac{3}{2}-1}$

Given:-

$V_1=1L$

$V_2=4L$

$T_1=320K$

$T_2=T\left(\text{say}\right)=?$

$\therefore \frac{T}{320}={\left(\frac{1}{4}\right)}^{\frac{1}{2}}$

$\Rightarrow T=160K$

Again, as we know that,

$\Delta U=n_{total}{C}_v\Delta T$

As $n_{total}=1+2=3$

$C_v=\frac{f_{av.}}{2}R=2R$

$\Delta T=160-320=-160K$

$\therefore \Delta U=3\times 2R\times (-160)$

$\Rightarrow \Delta U=-960R$

Hence $\Delta E$ of $\Delta U$ for the process is $-960R$.