Question

The internal energy of monoatomic ideal gas is $1.5nRT$. $1$ mole of helium is kept in a cylinder cross section $8.5c{m}^2$. The cylinder is closed by a light frictionless piston. The gas is heated in a process during which a total of $42J$ heat is given to the gas. If the temperature rises through. The distance moved by the piston is [Atmospheric pressure being $100kPa.]$

• A. $10cm$
• B. $20cm$
• C. $25cm$
• D. $30cm$

Answer 1

Answer: (B)

$20cm$

The change in internal energy of the gas is $\Delta U=1.5nR\left(\Delta T\right)$

$=1.5\left(1mol\right)\left(8.3J/mo{l}^{-1}K\right)\left(2K\right)=24.9J$

The heat is given to the gas $=42J$

The work done by the gas is $\Delta W=\Delta Q-\Delta U=42J-24.9J=17.1J$

If the distance moved by the piston is x, the work done is

$\delta W=\left(100kPa\right)\left(8.5c{m}^2\right)x$

Thus, $\left({10}^{5}N/m^2\right)\cdot 8.5\times 10-4m^2)x=17.1J$

or $x=0.2m=20cm$