A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting at a tempera

Question

A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting at a temperature of 300 K. The initial pressure of the ideal gas is atmospheric pressure, and the final pressure is four times the initial pressure.

Determine the following:
a. the change in the internal energy of the gas.
b. the work done by the gas.
c. the heat flow into or out of the gas.

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Eirian 4 years 2021-08-26T19:44:13+00:00 1 Answers 24 views 0

Answers ( )

  1. Nick
    0
    2021-08-26T19:46:02+00:00
    Reply

    Answer:

    a) 2250 J

    b) 0 J

    c) 2250 J

    Explanation:

    a) Since, the process is isochoric

    the change in internal energy

    \Delta U = n C_v(T_f-T_i)

    Here, n = 0.2 moles

    Cv = 12.5 J/mole.K

    We have to find T_f so we can use gas equation as

    \frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K

    So,  \Delta U= 0.2\times12.5(1200-300)\\=2250 J

    b) Since, the process is isochoric no work shall be done.

    c) By first law of thermodynamics we have

    \Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J

    Since, Q is positive 2250 J of heat will flow into the system.

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