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 cm^(2)`. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42J heat is given to the gas. if the temperature rise through `2^((0))C`, find the distance moved by the piston. atmoshphere pressure `=100 kPa.

Explanation

The change in internal energy of the gas is `DeltaU=1.5 nR(DeltaT)`
`=1.5(1 mol)(8.3 JK^(-1))(2 K)`
`24.9 J`.
time heat given to the gas `=42J`.
the work done by the gas is
`DeltaW=DeltaQ-DeltaU`
`=42 J-24.9 J=17.1 J`.
if the distance moved by the piston is x, the work done is
`DeltaW=(100 kPa)(8.5 cm^(2))x`.
Thus,
`(10^(5)Nm^(-2)) (8.5xx10^(-4)m^(-2))x=17.1 J`
or, `x=0.2m=20 cm`.