Question

Calculate the work done by system in an irreversible (single step) adiabatic expansion of $1$ mole of a polyatomic gas $(\lambda =4/3)$ from $400K$ and pressure $10atm$ to $1atm$.

Answer 1

Given:-

Initial temperature $\left(T_1\right)=400K$

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

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

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

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

For an adiabatic process,

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

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

$\Rightarrow \left(\frac{400}{T}\right)={\left(\frac{10}{1}\right)}^{1-\frac{1}{\gamma }}$

$\Rightarrow \frac{400}{T}={\left(10\right)}^{\frac{1}{4}}$

$\Rightarrow T=\frac{400}{1.77}=226K=T_2$

$\therefore \Delta T=(226-300)=-174K$

Therefore, for an addiabatic expansion,

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

$W=1\times \frac{R}{\gamma -1}\times (-174)$

$\Rightarrow W=3R\times (-174)=-522R$

Hence work done by the system will be $-522R$.