Question

One mole of a monatomic ideal gas undergoes the process $A\to B$ in the given $p-V$ diagram. Specific heat capacity in the process is

• A. $\frac{13R}{3}$
• B. $\frac{13R}{6}$
• C. $\frac{7R}{3}$
• D. $\frac{2R}{3}$

Answer 1

Answer: (B)

According to the question

Now, the temperature at point $A$ and $B$
By, $pV=nRT$ (from ideal gas equation) Temperature at point $A$,
$T_A=\frac{3p_0{V}_0}{R}$
Temperature at point $B$,
or, $T_B=\frac{30p_0{V}_0}{R}$
So, the temperature difference, $\Delta T=T_B-T_A$
$\therefore \Delta T=\frac{30p_0{V}_0}{R}-\frac{3p_0{V}_0}{R}$
$\Delta T=\frac{27p_0{V}_0}{R}$
Now, due to the adiabatic process, change in internal energy,
$\therefore \Delta U=n{C}_v\Delta T$
Or,$\Delta U=\frac{3}{2}R\Delta T...(i)$
Work done by the gas undergoes the process $A\to B$
$\therefore W=$ Area under $p-V$ graph
or$W=18p_0{V}_0$
or$W=\frac{2}{3}R\Delta T...(ii)$
Now, by first law of thermodynamics,
$Q=\Delta U+W$
By putting the values from Eqs. (i) and (ii) to above Eq. we get
$\therefore Q=\frac{3}{2}R\Delta T+\frac{2}{3}R\Delta T$
or $C\Delta T=\frac{13}{6}R\Delta T$
or $(\because Q=nC\Delta T,$ where $n=1$ mole $)$
$C=\frac{13}{6}R$
So, the specific heat capacity in the process $A\to B$ is $C=\frac{13}{6}R$