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^(@)C`, find the distance moved by the piston. atmospheric pressure `=100 kPa.`

Options

A. `10 cm`
B. `15 cm`
C. `20 cm`
D. `5 cm`

Answer

C

Explanation

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