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.

The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of h

1.	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.
Answer» 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`.

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