Question

Two mole of ideal diatomc gas ($C_{v,m}=\frac{5}{2}R$) at 300 K and 5 atm is expanded irreversibly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, w, H and U.

Answer 1

Given data is,

$P_1=5atm,P_2=2atm,P_{ext}=1$

$T_1=300K,C_v=\frac{5}{2}R,n=2$

For a Adiabatic Process

$w=n{C}_v(T_2-T_1)=-P_{ext}(\frac{nR{T}_2}{P_2}-\frac{nRT}{P_1})$

$w=2\times \frac{5}{2}R(T_2-300)=-1\times nR(\frac{T_2}{P_2}-\frac{T_1}{P_1})$

$\Rightarrow 5R(T_2-300)=-1\times 2\times R(\frac{T_2}{2}-\frac{300}{5})$

$\Rightarrow 5(T_2-300)=-2\left(\frac{5T_2-600}{10}\right)$

$\Rightarrow 5T_2-1500=-2\times 5\left(\frac{T_2-120}{10}\right)$

$\Rightarrow 5T_2-1500=-10\left(\frac{T_2-120}{10}\right)$

$\Rightarrow 5T_2-1500=-T_2+120$

$\Rightarrow 6T_2=1620$

$\Rightarrow T_2=270K$

$\Delta T=270-300=-30K$

$w=n{C}_v(T_2-T_1)$

$\Rightarrow 2\times \frac{5R}{2}\times (-30)$

$\Rightarrow 5\times 8.314\times -30=-1247.1J$

$q=0$

$\Delta U=w$

$\Delta H=\Delta U+nR\Delta T$

$\Rightarrow -1247.1+2\times 8.314\times -30$

$\Rightarrow -1745.94J$

So,

$q=0,w=-1247.1J,\Delta U=-1247.1J,\Delta H=-1745.94J.$