Question

One mole of a monoatomic ideal gas is taken through the cycle shown in figure.
$A\to B:$ Adiabatic expansion
$B\to C:$ Cooling at constant volume
$C\to D:$ Adiabatic compression.
$D\to A:$ Heating at constant volume
The pressure and temperature at $A,B,C$ and $D$ are denoted by $(P_A,T_A),(P_B,T_B),(P_C,T_C)$ and $(P_D,T_D)$ respectively. Given $T_A=1000\text{K},P_B=\frac{2}{3}{P}_A$ and $P_C=\frac{1}{3}{P}_A$, the work done by the gas in the process $A\to B$ is


[Take $R=8.31\text{J/mol K}$]

• A. $1869.75\text{J}$
• B. $970.23\text{J}$
• C. $1123\text{J}$
• D. $875.32\text{J}$

Answer 1

The correct option is A $1869.75\text{J}$

Let $\mu$ be the number of moles of an ideal gas. Given ideal gas is monoatomic. Thus, $\gamma =\frac{5}{3}$.
For adiabatic change, equation of state given in terms of $P$ and $T$ is given by
$\frac{T^{\gamma }}{P^{\gamma -1}}$ = constant
$\Rightarrow {\left(\frac{T_B}{T_A}\right)}^{\gamma }={\left(\frac{P_B}{P_A}\right)}^{\gamma -1}$
$\Rightarrow T_B=T_A{\left(\frac{P_B}{P_A}\right)}^{1-\frac{1}{\gamma }}$
From the data given in the question,
$T_B=1000\times {\left(\frac{2}{3}\right)}^{2/5}=850\text{K}$

Work done during an adiabatic process is given by
$W_{AB}=\frac{\mu R[T_A-T_B]}{\gamma -1}$
From the data given in the question,
$W_{AB}=\frac{1\times 8.31\times [1000-850]}{\left(5/3\right)-1}$
$\Rightarrow W_{AB}=\frac{3}{2}\times 8.31\times 150=1869.75\text{J}$
Thus, option (a) is the correct answer.