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) (2K )`
`= 24.9J`
The heat given to the `= 42 J`
The work done by the gas is
`Delta W = Delta Q- Delta U=42 J -24.9J =17.1J`
If the distance moved by the piston is x, the work done is
`Delta W=(100 kPa) (8.5 cm^(2))`
Thus, `(10^(5)N//m ^(2)) (8.5 xx 10 ^(4) m ^(2)) x = 17.1 J`
or, `x =0.2 m=20cm`