Question

The internal energy of monoatomic ideal gas is $1.5 \,nRT$. Two moles of helium is kept in a cylinder of cross-section $9.055 \,cm ^{2}$. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of $185.62\, J$ heat is given to the gas. If the temperature rises through $6^{\circ} C$, find the distance moved by the piston (in $cm$ ). Consider atmospheric pressure to be $100\, kPa$ and universal gas constant $=8.3\, J\, mol ^{-1} K ^{-1}$.

Answer 1

$\Delta U =1.5\, nR (\Delta T )=1.5(2)(8.3)(6)=149.4 \,J$
The heat given to the gas $=185.62 \,J$
The work done by the gas is
$\Delta W =\Delta Q -\Delta U =185.62-149.40$
$ \therefore \Delta W =36.22 \,J$
If the distance moved by the piston is $ x $, then work done is,
$ \Delta W = PA x $
$ \therefore 36.22=10^{5} \times 9.055 \times 10^{-4} \times( x ) \ldots . .[\text { From (i) }] $
$ \therefore x = \frac{36.22}{90.55}$
$=\frac{18.11 \times 2}{18.11 \times 5}=\frac{2}{5}$
$=0.4 \,m =40 \,cm .$