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10 July, 22:38

A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm to a final pressure of 1.00 atm in two ways: (a) reversibly, and (b) against a constant external pressure of 1.00 atm. Evaluate q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔStot in each case.

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  1. 11 July, 01:20
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    a) reversibly

    ΔU = 0

    q = 2740.16 J

    w = - 2740.16 J

    ΔH = 0

    ΔS (total) = 0

    ΔS (sys) = 9.13 J/K

    ΔS (surr) = - 9.13 J/K

    b) against a constant external pressure of 1.00 atm

    ΔU = 0

    w = - 1.66 kJ

    q = 1.66 kJ

    ΔH = 0

    ΔS (sys) = 9.13 J/K

    ΔS (surr) = - 5.543 J/K

    ΔS (total) = 3.587 J/K

    Explanation:

    Step 1: Data given:

    Number of moles = 1.00 mol

    Temperature = 27.00 °C = 300 Kelvin

    Initial pressure = 3.00 atm

    Final pressure = 1.00 atm

    The gas constant = 8.31 J/mol*K

    (a) reversibly

    Step 2: Calculate work done

    For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

    Since the temperature remains constant:

    ΔU = 0

    ΔU = q + w

    q = - w

    w = - nRT ln (Pi/Pf)

    ⇒ with n = the number of moles of perfect gas = 1.00 mol

    ⇒ with R = the gas constant = 8.314 J/mol*K

    ⇒ with T = the temperature = 300 Kelvin

    ⇒ with Pi = the initial pressure = 3.00 atm

    ⇒ with Pf = the final pressure = 1.00 atm

    w = - 1*8.314 * 300 * ln (3)

    w = - 2740.16 J

    q = - w

    q = 2740.16 J

    Step 3: Calculate change in enthalpy

    Since there is no change in energy, ΔH = 0

    Step 4: Calculate ΔS

    for an isothermal process

    ΔS (total) = ΔS (sys) + ΔS (surr)

    ΔS (sys) = - ΔS (surr)

    ΔS (sys) = n*R*ln (pi/pf)

    ΔS (sys) = 1.00 * 8.314 * ln (3)

    ΔS (sys) = 9.13 J/K

    ΔS (surr) = - 9.13 J/K

    ΔS (total) = ΔS (sys) + ΔS (surr) = 0

    (b) against a constant external pressure of 1.00 atm

    Step 1: Calculate the work done

    w = - Pext*ΔV

    w = - Pext * (Vf - Vi)

    ⇒ with Vf = the final volume

    ⇒ with Vi = the initial volume

    We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

    V = (n*R*T) / P

    Initial volume = (n*R*T) / Pi

    ⇒ Vi = (1*0.08206 * 300) / 3

    ⇒ Vi = 8.206 L

    Final volume = (n*R*T) / Pf

    ⇒ Vf = (1*0.08206 * 300) / 1

    ⇒ Vf = 24.618 L

    The work done w = - Pext * (Vf - Vi)

    w = - 1.00 * (24.618 - 8.206)

    w = - 16.412 atm*L

    w = - 16.412 * (101325/1atm*L) * (1kJ/1000J)

    w = - 1662.9 J = - 1.66 kJ

    Step 2: Calculate the change in internal energy

    ΔU = 0

    q = - w

    q = 1.66 kJ

    ΔH = 0 because there is no change in energy

    Step 3: Calculate ΔS

    ΔS (sys) = n*R*ln (3)

    ΔS (sys) = 1.00 * 8.314 * ln (3)

    ΔS (sys) = 9.13 J/K

    ΔS (surr) = - q/T

    ΔS (surr) = - 1662.9J/300K

    ΔS (surr) = - 5.543 J/K

    ΔS (total) = ΔS (surr) + ΔS (sys) = - 5.543 J/K + 9.13 J/K = 3.587 J/K
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