Question

The work done by the system in an irreversible (single step) adiabatic expansion of $1$ mole of a polyatomic gas $(\gamma =1.33)$ from $300$K and pressure $10$ atm to $1$ atm is positive.

• A. True
• B. False

Answer 1

The correct option is B False

Solution:- (B) False

Given:-

Initial temperature $\left(T_1\right)=300K$

Final temperature $\left(T_2\right)=T\left(\text{say}\right)=?$

Initial pressure $\left(P_1\right)=10atm$

Final pressure $\left(P_2\right)=1atm$

$\gamma =1.33=\frac{4}{3}$

For an adiabatic process,

$\frac{T^{\gamma }}{P^{\gamma -1}}=\text{constant}$

$\therefore {\left(\frac{T_1}{T_2}\right)}^{\gamma }={\left(\frac{P_1}{P_2}\right)}^{\gamma -1}$

$\Rightarrow \left(\frac{300}{T}\right)={\left(\frac{10}{1}\right)}^{1-\frac{1}{\gamma }}$

$\Rightarrow \frac{300}{T}={\left(10\right)}^{\frac{1}{4}}$

$\Rightarrow T=\frac{300}{1.77}=168.72K=T_2$

$\therefore \Delta T=(168.72-300)=-131.28K$

Therefore, for an addiabatic expansion,

$W=n{C}_V\Delta T$

$W=1\times \frac{R}{\gamma -1}\times (-131.28)$

$\Rightarrow W=3R\times (-131.28)=-393.84R$

Hence work done by the system will be $-ve$.

Hence the given statement is false.