Work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7^ . T - Physics MCQ AIEEE 2006 2026 | ExamTest.live

mediumMCQAIEEE 20062026Physics

1 attempts0% success rate1 mark

Work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7^{\circ} C . The gas is: (R = 8.3 J mol^{- 1} K^{- 1})

Solution & Step-by-step Explanation

For an adiabatic process, W = Δ U = n C_{v} Δ T .Given W = 146 \times 1 0^{3} J, n = 1000 moles (1 kilo mole), and Δ T = 7 K . 146000 = 1000 \times C_{v} \times 7 146 = 7 C_{v} ⟹ C_{v} = \frac{146}{7} \approx 20.86 J/mol K We know C_{v} = \frac{f}{2} R . 20.86 = \frac{f}{2} (8.3) f = \frac{2 \times 20.86}{8.3} \approx \frac{41.72}{8.3} \approx 5 A degrees of freedom value of 5 corresponds to a diatomic gas.

Try it yourself before checking the explanation above.

Work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7^{\circ} C . The gas is: (R = 8.3 J mol^{- 1} K^{- 1})

monoatomic

diatomic

triatomic

a mixture of monoatomic and diatomic

Practice with timed mock tests and track your performance across Physics.