Question

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

• A. $20\text{cm}$
• B. $30\text{cm}$
• C. $40\text{cm}$
• D. $50\text{cm}$

Answer 1

The correct option is A $20\text{cm}$

Given :
$n=1,A=8.5\times {10}^{-4}{\text{m}}^2$
$\Delta Q=42\text{J},\Delta T=2^{\circ }C$

Change in internal energy $=1.5nR\Delta T$
$\Delta U=1.5\times 1\times 8.31\times 2=24.93\text{J}$
$\Delta Q=\Delta U+\Delta W$
$42=24.93+\Delta W$
$\Delta W=17.07\text{J}$
Now at any instant, F.B.D. of piston will be

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