Question

A monoatomic ideal gas has an internal energy $1.5nRT$ . A cylinder has cross-section $9.055c{m}^2$ . It contains $2moles$ of $He$ gas. 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

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