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InterviewSolution
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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.`A. `10 cm`B. `15 cm`C. `20 cm`D. `5 cm` |
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Answer» Correct Answer - C 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`. |
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