Question

The internal energy of a monoatomic ideal gas is $1.5nRT$. One mole of helium is kept in a cylinder of cross-section $A=8.5c{m}^2$. The cylinder is closed by a light frictionless piston. The gas is slowly heated in a process during which a total of $42J$ heat is given to the gas. If the temperature rises through $2^{\circ }C$, find the displacement $(dx$ ) (in $cm$) of the piston. Atmospheric pressure $=100kPa$.

Answer 1

Answer: 20.00

Given that $n=1,A=8.5\times 10^{-4}{m}^2$
$\Delta Q=42J,\Delta T=2^{\circ }C$
Change in internal energy $=1.5nR\Delta T$
$\Delta U=1.5\times 1\times 8.31\times 2=24.93J$
$\Delta Q=\Delta U+\Delta W$
$42=24.93+\Delta W$
$\Delta W=17.07J$
Now at any instant $F.B.D$. of piston

As piston is moved slowly, so it means change in velocity in a given time is $0.$ Hence acceleration is $0$ .
Since $F_{\text{net }}=0$
$\Rightarrow P_{0}A-PA=0$
$\Rightarrow P=P_0$
$\therefore \Delta W=$ Work done by gas
$=Fx\cos 0^{\circ }=\left(P_{0}A\right)(\Delta x)(1)$
$17.07=\left(100\times 10^3\times 8.5\times 10^{-4}\right)\Delta x$
$\Delta x=\frac{17.07}{85}=0.2m=20cm$