Question

The internal energy of monoatomic ideal gas is $1.5nRT$. Two moles of helium is kept in a cylinder of cross-section $9.055c{m}^2$. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of $185.62J$ 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 $100kPa$ and universal gas constant $=8.3Jmo{l}^{-1}{K}^{-1}$.

Answer 1

Answer: 40

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