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One mole of an ideal gas (CV= 1.5R)t temperature 500 k is compressed from 1.0 atm 2.0 atm by an reversible isothermal path. Subsequently it is expanded back to 1.0atm by reversible adiabatic path. The volume of final state in litre is ?
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One mole of an ideal gas (CV= 1.5R)t temperature 500 k is compressed f...
Answer:

Given:

  • Number of moles (n) = 1 mol

  • Cv = 1.5R

  • T = 500 K

  • P1 = 1.0 atm

  • P2 = 2.0 atm

  • Path 1: Reversible isothermal process

  • Path 2: Reversible adiabatic process



Calculations:

Step 1: Finding the initial volume (V1)

  • Using ideal gas equation, PV = nRT

  • V1 = nRT1/P1 = (1 mol x 8.314 J/K/mol x 500 K) / (1.0 atm x 101.325 kPa/atm) = 41.24 L



Step 2: Finding the final volume (Vf)

  • Path 1: Isothermal compression


    • As the process is isothermal, temperature remains constant at T = 500 K

    • Using the ideal gas equation, P1V1 = P2V2

    • V2 = P1V1/P2 = (1.0 atm x 41.24 L) / (2.0 atm) = 20.62 L


  • Path 2: Adiabatic expansion


    • As the process is adiabatic, there is no heat exchange with the surroundings (Q = 0)

    • Using the adiabatic relation, PV^(γ) = constant

    • For an ideal gas, γ = Cp/Cv = (Cv + R)/Cv = (1.5R + R)/1.5R = 5/3

    • At the final state, P3Vf^(5/3) = P2V2^(5/3)

    • Vf = [P2V2^(5/3)] / P3^(5/3)

    • As the process is reversible, P3 = P1 = 1.0 atm

    • Vf = [2.0 atm x (20.62 L)^(5/3)] / (1.0 atm)^(5/3) = 31.71 L




Step 3: Finding the final volume at P = 1.0 atm (Vf')

  • Using the ideal gas equation, P3Vf' = P2Vf

  • Vf' = P2Vf/P3 = (2.0 atm x 31.71 L) / (1.0 atm) = 63.42 L



Final Answer:
The final volume of the gas at P = 1.0 atm is 63.42 L.
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One mole of an ideal gas (CV= 1.5R)t temperature 500 k is compressed f...
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One mole of an ideal gas (CV= 1.5R)t temperature 500 k is compressed from 1.0 atm 2.0 atm by an reversible isothermal path. Subsequently it is expanded back to 1.0atm by reversible adiabatic path. The volume of final state in litre is ?
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