A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 at

Question

A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J

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Thu Thủy 3 years 2021-07-15T12:28:45+00:00 1 Answers 6 views 0

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  1. MichaelMet
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    2021-07-15T12:30:07+00:00
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    Answer:

    a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K

    b) the change in internal energy is -18279.78 J

    c) heat lost by the gas is zero or 0

    d) the work done on the gas is -18279.78 J

    Explanation:

    Given the data in the question;

    P_i = 1 atm = 101325 pascal

    P_f = ?

    V_i = 0.1550 m³

    V_f = 0.742 m³

    we know that for an adiabatic process  γ = 1.4

    P_iV_i^Y = P_fV_f^Y

    P_f = P_i[tex]([/tex] V_i / V_f )^Y

    we substitute

    P_f = 1 × ( 0.1550  / 0.742  )^{1.4

    = ( 0.2088948787 )^{1.4

    = 0.11166 atm

    a) the initial and final temperatures

    Initial temperature

    T_i = P_iV_i / nR

    given that n = 3.10 mol

    = ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )

    = 15705.375 / 25.7734

    T_i  = 609.4 K

    Final temperature

    T_f = P_fV_f / nR

    = ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )

    = 8394.95 / 25.7734

    = 325.7 K

    Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K

    b) the change in internal energy

    ΔE_{int = nC_vΔT

    here, C_v = ( 5/2 )R

    ΔE_{int = 3.10 × ( 5/2 )8.314 × ( 325.7 – 609.4 )

    = -18279.78 J

    Therefore, the change in internal energy is -18279.78 J

    c) the heat lost by the gas

    Since its an adiabatic process,

    Q = 0

    Therefore, heat lost by the gas is zero or 0

    d)  the work done on the gas

    W = ΔE_{int – Q

    = -18279.78 J – 0

    W = -18279.78 J

    Therefore, the work done on the gas is -18279.78 J

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