A 100cm3 box contains helium at a pressure of 2.40atm and a temperature of 120 d

Question

A 100cm3 box contains helium at a pressure of 2.40atm and a temperature of 120 degree C. It is placed in thermal contact with a 200cm3 box containing argon at a pressure of 4.40atm and a temperature of 410 degree C. What is the final temperature? Correct Significant Figures Feedback: Your answer 316.75 degree C was either rounded differently or used a different number of significant figures than required for this part. What is the final pressure in each box?

Explanation / Answer

Thermal energy of an ideal gas is given by: U = n·Cv·T (n number of moles, Cv molar heat capacity at constant volume , T absolute temperature) The molar heat capacity of monatomic ideal gases like helium and argon is: Cv = (3/2)·R The number of moles of each gas can be found from ideal gas law: p·V = n·R·T => n = p·V/(R·T) -helium n1 = p1·V1/(R·T1) = 1.6atm · 0.12L / ( 0.08205746atmL/molK · (120+273.15)K) = 5.95×10?³mol -argon n2 = p2·V2/(R·T2) = 4.0atm · 0.2L / ( 0.08205746atmL/molK · (380+273.15)K) = 14.93×10?³mol So the initial energies are: -helium U1 = (3/2)·n1·R·T1 = (3/2) · 5.95×10?³mol · 8.314472J/molK · (120+273.15)K = 29.18J -argon U2 = (3/2)·n2·R·T2 = (3/2) · 14.93×10?³mol · 8.314472J/molK · (380+273.15)K = 121.59J b. Assuming threre is only energy exchanged between the boxes, the total internal energy is conserved. So the sum of the final thermal energies equals the sum of the initial energies: U1' + U2' = U1 + U2 On the other hand both boxes have same final temperature final T' at equilibrium: U1' = (3/2)·n1·R·T' U2' = (3/2)·n2·R·T' hence: U1'/n1 = U2'/n2 U2' = (n2/n1) · U1' substitute to energy balance U1' + (n2/n1) · U1' = U1 + U2 => U1' = (U1 + U2) / (1 + (n2/n1)) = (29.18J + 121.59J) / (1 + (14.93×10?³mol/5.95×10?³mol) ) = 42.98J => U2' = U1 + U2 - U1' = 29.18J + 121.59J - 42.98J = 107.79J c. The heat energy transferred is equal to the absolute value of thermal energy change in each box: ?U1 = U1' - U1 = 42.98J - 29.18 = 13.8J ?U1 > 0 , so the thermal energy of the helium has risen, while argon's thermal energy has dropped by the same value. That means heat is transferred from argon to helium. d. U1' + U2' = U1 + U2 (3/2)·n1·R·T' + (3/2)·n2·R·T' = U1 + U2 => T' = (U1 + U2) / ((3/2)·R·(n1 + n2)) = (29.18J + 121.59J) / ((3/2)·8.314472J/molK·(5.95×10?³mol + 14.93×10?³mol)) = 579K = 306°C e. since n and V is constant in each box p/T = n·R/V = constant p' = p·(T'/T) => p1' = p1·(T'/T1) = 1.6atm · (579K/ (120+273.15)K) = 2.356atm p2' = p2·(T'/T2) = 4.0atm · (579K/ (380+273.15)K) = 3.546atm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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