Question

One mole of an ideal monoatomic gas expands reversibly and adiabatically from a volume of $x$ litre to 14 litre at ${27}^{\circ }C$. Then the value of $x$ will be:[Final temperature $=189\text{K}$ and $C_v=\frac{3}{2}\text{R}$ ]

Answer 1

$T_i=300K,T_f=189K.$
$V_i=xL,V_f=14L$
In an adiabatic, reversible process:
$T{V}^{\gamma -1}=constant(and\gamma =\frac{5}{3})$
$\therefore 300\times (x{)}^{\frac{2}{3}}=189\times (14{)}^{\frac{2}{3}}$
$\therefore x=14\times {\left(\frac{189}{300}\right)}^{\frac{3}{2}}=14\times {\left(\frac{63}{100}\right)}^{\frac{3}{2}}$
$\therefore x=7$