Question

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

Answer 1

Given that $n =1, A =8.5 \times 10^{-4} m ^{2}$
$\Delta Q =42 \,J , \Delta T =2^{\circ} C$
Change in internal energy $=1.5 \,nR \Delta T$
$\Delta U =1.5 \times 1 \times 8.31 \times 2=24.93 \,J$
$\Delta Q =\Delta U +\Delta W $
$42 =24.93+\Delta W$
$\Delta W =17.07 \,J$
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 $
$\Delta x =\frac{17.07}{85}=0.2 \,m =20 \,cm$