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4 October, 08:47

One mole of an ideal gas, CP=3.5R, is compressed adiabatically in a piston/cylinder device from 2 bar and 25 oC to 7 bar. The process is irreversible and requires 35% more work than a reversible, adiabatic compression from the same initial state to the same final pressure. What is theentropy change of the gas?

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  1. 4 October, 10:05
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    the entropy change of the gas is ΔS = 2.913 J/K

    Explanation:

    starting from the first law o thermodynamics for an adiabatic reversible process

    ΔU = Q - W

    where

    ΔU = change in internal energy

    Q = heat flow = 0 (adiabatic)

    W = work done by the gas

    then

    -W = ΔU

    also we know that the ideal compression work Wcom = - W, then Wcom = ΔU. But also for an ideal gas

    ΔU = n*cv * (T final - T initial)

    where

    n=moles of gas

    cv = specific heat capacity at constant volume

    T final = T₂ = final temperature of the gas

    T initial = T₁ = initial temperature of the gas

    and also from an ideal gas

    cp - cv = R → cv = 7/2*R - R = 5/2*R

    therefore

    W com = ΔU = n*cv * (T final - T initial)

    for an ideal gas under a reversible adiabatic process ΔS=0 and

    ΔS = cp*ln (T₂/T₁) - R * ln (P₂/P₁) = 0

    therefore

    T₂ = T₁ * (P₂/P₁) ^ (R/cp) = T₁ * (P₂/P₁) ^ (R / (7/2R)) = T₁ * (P₂/P₁) ^ (2/7)

    replacing values T₁=25°C = 298 K

    T₂ = T₁ * (P₂/P₁) ^ (2/7) = 298 K * (7 bar/2 bar) ^ (2/7) = 426.25 K

    then

    W com = ΔU = n*cv * (T₂ - T₁)

    and the real compression work is W real = 1.35*Wcom, then

    W real = ΔU

    W real = 1.35*Wcom = n*cv * (T₃ - T₁)

    T₃ = 1.35*Wcom/n*cv + T₁ = 1.35 * (T₂ - T₁) + T₁ = 1.35*T₂ - 0.35*T₁ = 1.35*426.25 K - 0.35 * 298 K = 471.14 K

    T₃ = 471.14 K

    where

    T real = T₃

    then the entropy change will be

    ΔS = cp*ln (T₃/T₁) - R * ln (P₂/P₁) = 7/2 * 8.314 J/mol K * ln (471.14 K / 298 K) - 8.314 J/mol K * ln (7 bar / 2 bar) = 2.913 J/K

    ΔS = 2.913 J/K
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