Question

Three moles of an ideal monotomic gas performs a cycle $ABCDA$ as shown in the figure. The temperatures of the gas at the states $A,B,C$ and $D$ are $400K,800K,2400K$ and $1200K$, respectively. The work done by the gas during this cycle is ( $R$ is universal gas constant)

• A. 1200 R
• B. 3600 R
• C. 2400 R
• D. 2000 R

Answer 1

Answer: (C)

According to the question,

Given,
number of moles of an ideal monoatomic gas, $n=3$,
temperature of the gas of state $A,T_A=400K$,
temperature of the gas at state $B,T_B=800K$,
temperature of the gas at state $C,T_C=2400K$,
and temperature of the gas at state $D,T_D=1200K$,
Now, work done by the gas during this cycle is, $W=nR\Delta T$
$(\therefore$ Work done in a isothermal process)
So, $W_{\text{net }}=W_{A\to B}+W_{B\to C}+W_{C\to D}+W_{D\to A}$
$W_{\text{net }}=nR\left[\left(T_B-T_A\right)+\left(T_C-T_B\right)+\left(T_D-T_C\right)-\left(T_D-T_A\right)\right]$
$W_{\text{net }}=3R[0+(2400-800)+0-(1200-400)]$
$W_{\text{net }}=3R\times [1600-800]$
$W_{\text{net }}=3R\times 800,W_{\text{net }}=2400R$
So, the work done by the gas during,
$A\to B\to C\to D$ cycle is $W_{\text{net }}=2400R$