A mass of 2.4 kg of air at 150 kPa and 12°C is contained in a gas - tight, frictionless piston - cylinder device. The air is then compressed to a final pressure of 600 kPa. During this process, heat is transferred from the air in such a way that the temperature inside the cylinder remains constant. Calculate the work input during the process.

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SSC JE Mechanical 4 Dec 2023 Official Paper - II
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  1. -272 kJ
  2. 272 kJ
  3. -11 kJ
  4. 11 kJ

Answer (Detailed Solution Below)

Option 1 : -272 kJ
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Detailed Solution

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Concept:

During isothermal compression, temperature remains constant and the work done by the gas is given by

\(dW = {P_1}{V_1}\ln \frac{{{V_2}}}{{{V_1}}} = {P_1}{V_1}\ln \frac{{{P_1}}}{{{P_2}}}\) = \(mRT_1 \ln \frac{{{P_1}}}{{{P_2}}}\)

Calculation:

Given:

P1 = 150 kPa, T = 12°C = 285 K, P2 = 600 kPa, m = 2.4 kg, R = 0.287 kJ/Kg-K

The work done by the gas is

W = \(mRT_1 \ln \frac{{{P_1}}}{{{P_2}}}\) = \( 2.4 \times {0.287} \times 285~\times\ln \frac{{150}}{{600}} = -272.14\;kJ\;\)

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