Question

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

• A. $w=-1246.1J,q<0,\Delta H=-1745.9J,\Delta U=0$
• B. $w=1745J,q=0,\Delta H=1247J,\Delta U=0$
• C. $w=-1246J,q=0,\Delta H=-1746J,\Delta U=-1247J$
• D. $w=-1745.6J,q>0,\Delta H=-1245J,\Delta U=-1245J$

Answer 1

The correct option is C $w=-1246J,q=0,\Delta H=-1746J,\Delta U=-1247J$

For Adiabatic irreversible process
$n{C}_v(T_2-T_1)=-p_{ext}nR[\frac{T_2}{p_2}-\frac{T_1}{P_1}]$ ...(1)
Where, $C_v=\frac{5}{2}R$
$T_1=300K$
$P_{ext}=1atm{P}_2=2atm$
$P_1=5atm$
$2\times \frac{5}{2}R(T_2-300)=-1\times 2R[\frac{T_2}{2}-\frac{300}{5}]$
$=5(T_2-300)=-(T_2-120)$
$=5T_2-1500=-T_2+120$
$\Rightarrow T_2=270k$
Also we know,
$\Delta U=q+w=w\because q=0$
The workdone in an adiabatic process= $w=n{C}_v△T$
$=2\times \frac{5}{2}\times 8.314(270-300)$
$=-1247.1J$
$\Delta U=w=-1247.1J$
$\Delta H=△U+nR\Delta T$
$=-1247.1+2\times 8.314(270-300)$
$\Rightarrow \Delta H=-1745.9J$