Question

The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross section 8.5$c{m}^2$. The cylinder is closed by a light frictionless piston. The gas is heated slowly 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 distance moved by the position. Atmospheric pressure = 100 kPa

• A. 2.5 cm
• B. 20 cm
• C. 20 mm
• D. 25 mm

Answer 1

The correct option is B

20 cm

$\Delta U=1.5nR\left(\Delta T\right)$
$=1.5\left(1mol\right)\left(8.3J{K}^{-1}mo{l}^{-1}\right)\left(2K\right)$
$=24.9J.$
The heat given to the gas $=42$ J.
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
$\Delta W=\left(100kPa\right)\left(8.5c{m}^2\right)xJ$
Thus,
$\left(100N{m}^{-2}\right)(8.5\times {10}^{-1}{m}^2)x=17.1J$

or,

$x=0.2m=20cm.$