A 80.0cm3 box contains helium at a pressure of 2.30atm and a temperature of 90.0

Question

A 80.0cm3 box contains helium at a pressure of 2.30atm and a temperature of 90.0?C. It is placed in thermal contact with a 220cm3 box containing argon at a pressure of 4.30atm and a temperature of 380?C.

A-) What is the initial thermal energy of each gas? Enter your answers numerically separated by a comma.

B-) What is the final thermal energy of each gas? Enter your answers numerically separated by a comma.

C-) How much heat energy is transferred, and in which direction?

D-) From Ar to He OR From He to Ar?

E-) What is the final temperature?

F-) What is the final pressure in each box? Enter your answers numerically separated by a comma.

Explanation / Answer

Eth=3/2nRT
= 3/2 *P*V.............(assuming ideal gas behaviour for both gases, PV = nRT)

a) Initial thermal energies are,

Helium = 3/2 * 2.3 * (80*10^-3) = .276 atm-lit ~ 28 joules

Argon = 3/2* 4.3*220*10^-3 = 1.42 atm-lit ~ 144 Joules

b) Finally, the temp of both gases will become same, volume and no. of moles remaining constant.

Thus, P/T is constant

EHe=nHe/(nHe+nAr) * (Etotal)

Etotal initial = Etotal final = .276+1.42 = 1.696 atm lit = 171.847 Joules

E He = (2.3*80/(90+273)) / (2.3*80/(90+273) + 4.3*220/(380+273)) * 171.847 = 44.54 Joules

E Ar = (4.3*220/(380+273)) / (2.3*80/(90+273) + 4.3*220/(380+273)) * 171.847 = 127.30 Joules

c & d) Heat is transferred from Ar to He as Ar is at higher temp. Total heat transferred is (EAr i - EAr f)

=143.88-127.30 = 16.58 Joules

e) 44.54 Joules = .44 atm-lit = 3/2*PV

P = .44*2/(3*80*10^-3) = 3.67 atm

P1/T1 = P2/T2

=> T2 = 3.67/2.3*(90+273) = 579.22 K = 306.22 deg C

f) PHe = 3.67 atm ( calculated above)

PAr = ?

P1/T1 = P2/T2

P2 = 4.3 * (306.22+273)/(380+273) = 3.81 atm

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