Consider 2 moles of an ideal diatomic gas at 300 K and atmospheric pressure P =

Question

Consider 2 moles of an ideal diatomic gas at 300 K and atmospheric pressure P = 1 x 10^5 N/m^2. Assume the Ideal Gas Law holds (PV = NRT) with N = the number of moles (2) and the universal gas constant R = 8.314 J / mol K

a. Using the equipartition theorem, what is the total internal energy E_int(300 K)

b. What is the molar specfic heat at constant volume c_v = 1 / N dE_int/dT

c. If Q = 3000 J of energy is added to the gas as heat with the volume kept the same, what is the new temperature T? What is the new pressure? What is the total internal energy, E_int(T).

d. The gas is allowed to expand isothermally (heat is added to keep the temperature constant at T) until it is twice the volume. What is the total internal energy? How much work was done in this step? How much heat was added?

Explanation / Answer

part a:

total internal energy=(3/2)*N*R*T

=1.5*2*8.314*300=7482.6 J

part b:

molar specific heat at constant volume=(5/2)*R=20.785 J/(mol.K)

part c:

as volume is unchanged, change in internal energy=heat added

=> number of moles*specific heat at constant voume*change in temperature=3000

==>2*20.785*(T-300)=3000

==>T=372.17 K

as voluem does not change, pressure/temperature remains constant

==>10^5 /300=pressure/372.17

==>pressure=1.2406*10^5 N/m^2

total internal energy=7482.6+3000=10482.6 J

part d:

as temperature remains constant, change in internal energy=0

so internal energy=10482.6 J

work done=n*R*T*ln(final volume/initial volume)

=2*8.314*372.17*ln(2)

=4289.5 J

so heat added=work done=4289.5 J

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.