Question

The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement is incorrect?

• A. $T_3>T_1$
• B. $T_1=T_2$
• C. $\Delta U_{\text{isothermal}}>\Delta U_{\text{adiabatic}}$
• D. $W_{\text{isothermal}}>w_{\text{adiabatic}}$

Answer 1

Answer: (A)

Use ideal gas equation $PV=nRT$ and consider,
Work done=area under P−V curve.
Step 1: Calculate the relation between $T_{1,}{T}_{2}and{T}_3$ ​ Since temperature is constant in any isothermal process, $T_1=T_2$ . From ideal gas equation $T_2=\frac{P_2{V}_2}{nR}and{T}_3=\frac{P_3{V}_3}{nR}$ ​
​ From the graph we can see $P_2>P_3$ and $T_2>T_3$ ​
So, relation between temperatures is: $T_1=T_2>T_3$ ​
Step 2: Compare the work done between the two processes. The work done is the area under the P−V curve. Since the isothermal curve is above the adiabatic curve at all times, the area under the isothermal curve is greater.
$⟹W$ isothermal $>W$ adiabatic
Step 3: Calculate $\Delta U$ for both the processes.
We know that $\Delta U=n{C}_V\Delta T$
$⟹\Delta U$ isothermal $=n{C}_V\left(T_2-T_1\right)=0$ (since, $T_1=T_2)$
and for adiabatic process we have
$⟹\Delta U$ adiabatic $={}^n{C}_V\left(T_3-T_1\right)<0,($ Since $T_1>T_3)$
Thus, $\Delta U$ isothermal $>\Delta U$ adiabatic