Question

n moles of an ideal gas undergoes a process $A\to B$ as shown in the figure. the maximum temperature of the gas during the process will be :

• A. $\frac{9P^0{V}^0}{4nR}$
• B. $\frac{9P^0{V}^0}{2nR}$
• C. $\frac{9P^0{V}^0}{nR}$
• D. $n=\frac{C_p}{C_v}$

Answer 1

The correct option is A $\frac{9P^0{V}^0}{4nR}$

The equation of line AB is P=$\frac{-P^0}{V^0}V+3P_0$

The point where PV will be maximum , temperature will be also maximum because for an ideal gas PV=nRT

So For finding that point we will do $\frac{d\left(PV\right)}{dV}=0$ which gives

$2\frac{-P_0}{V_0}V+3P_0=0$ which gives V=$\frac{3V_0}{2}$ and putting this in equation we get P=$\frac{3P_0}{2}$

So for Max Temp T${}_{max}$ , $\frac{9P_0{V}_0}{4}=nR{T}_{max}$

Which gives T${}_{max}=\frac{9P_0{V}_0}{4nR}$