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.`

Explanation

The change in internal energy of the gas is
`Delta U = 1.5 nR (Delta T)`
`= 1.5 (1 mol) (8.3 J//mol -K) (2 K) = 24. 9J`
The heat given to the gas `= 42 J`
The work done by the gas is
`Delta W = Delta Q - Delta U = 42 J - 24.9 J = 17.1 J`
If the distance moved by the piston is `x`, the work done is
`Delta W = (100 k Pa) (8.5 cm^(2)) x`
Thus, `(10^(5) N//m^(2)) *8.5 xx 10^(-4) m^(2)) x = 17.1 J`
or `x = 0.2 m = 20 cm`