Problem 20.51 Part A A 80.0 cm box contains helium at a pressure of 1.70 atm and

Question

Problem 20.51 Part A A 80.0 cm box contains helium at a pressure of 1.70 atm and a temperature of 80.0 °C. It is placed in thermal contact with a 190 cm3 box containing argon at a pressure of 3.70 atm and a temperature of 390 °C. What is the initial thermal energy of each gas? Enter your answers numerically separated by a comma. E(He).E(Ar)- Submit My Answers Give Up Incorrect; Try Again; 6 attempts remaining; no points deducted Your answer does not have the correct number of comma-separated terms Part B What is the final thermal energy of each gas? Enter your answers numerically separated by a comma. EfHe)E(Ar) Submit My Answers Give Up

Explanation / Answer

(A) Applying gas equation,

P V = n R T

(1.70 x 101325) (80 x 10^-6) = nHe (8.314) (273 + 80)

nHe = 4.695 x 10^-3 mol

(3.70 x 101325) (190 x 10^-6) = nAr (8.314) (273 + 390)

nAr = 12.92 x 10^-3 mol

E = 3 n R T / 2

E_He = (3/2) (4.695 x 10^-3) (8.314) (273 + 80)

= 20.7 J

E_Ar = 106.8 J

Ans: 20.7 , 106.8

(B) now total energy will be in the ratio of mol as temp is same for both.

Ef_He + Ef_Ar = 20.7 + 106.8 = 127.5 .... (i)

and Ef_He / Ef_Ar = 4.695 / 12.92

12.92 Ef_He - 4.695 Ef_Ar = 0 ...... (ii)

Solving both equations,

EF_He = 33.9 J

Ef _Ar = 93.5 J

(C) 33.9 - 20.7 = 13.2 J

(D) Ar to He

(E) 93.5 = (3/2) (12.92 x10^-3) (8.314) (T)

T = 580 K Or 307 deg C

(f) 1.70 / (273 + 80) = PfHe / 580

Pfhe = 2.79 atm Or 2.83 x 10^5 Pa

3.70 / (273 + 390) =PfAr / 580

PfAr = 3.24 atm Or 3.28 x 10^5 Pa

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