Question

One mole of an ideal gas $⟮C_v,m=\frac{5}{2}R⟯$ at 300 K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas is:

• A. $270$ K
• B. $273$ K
• C. $248.5$ K
• D. $200$ K

Answer 1

The correct option is C $248.5$ K

Solution:- (C) $248.5K$

Given that:-

$P_{ext.}=2atm$

$P_1=5atm$

$P_2=2atm$

$T_1=300K$

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

$n=1$

$C_V=\frac{5}{2}R$

For an adiabatic process,

$W=n{C}_{V}dT=-P\Delta V$

$\therefore n{C}_V(T_2-T_1)=-P_{ext}(V_2-V_1)$

$\Rightarrow 1\times \frac{5}{2}R(T-300)=-2(\frac{nR{T}_2}{P_2}-\frac{nR{T}_1}{P_1})$

$\Rightarrow \frac{5}{2}R(T-300)=-2R(\frac{T}{2}-\frac{300}{5})$

$\Rightarrow 5T-1500=-2T+240$

$\Rightarrow 7T=1740\Rightarrow T=248.57K$

$\therefore T_2=T=248.57K$