A 120cm^3 box contains helium at a pressure of 1.50atm and a temperature of 90.0

Question

A 120cm^3 box contains helium at a pressure of 1.50atm and a temperature of 90.0^circ C. It is placed in thermal contact with a 210cm^3 box containing argon at a pressure of 3.70atm and a temperature of 420^circ C. Part A What is the initial thermal energy of each gas? Part B What is the final thermal energy of each gas? Part C How much heat energy is transferred, and in which direction? Part D Part E What is the final temperature? Part F What is the final pressure in each box?

Explanation / Answer

a. 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) = 1.5atm · 0.12L / ( 0.08205746atmL/molK · (90+273.15)K) = in mol -argon n2 = p2·V2/(R·T2) = 3.70atm · 0.2L / ( 0.08205746atmL/molK · (420+273.15)K) = in mol So the initial energies are: -helium U1 = (3/2)·n1·R·T1 -argon U2 = (3/2)·n2·R·T2 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)) => U2' = U1 + U2 - U1' = c. The heat energy transferred is equal to the absolute value of thermal energy change in each box: ?U1 = U1' - U1 =in J ?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)) = e. since n and V is constant in each box p/T = n·R/V = constant p' = p·(T'/T) => p1' = p1·(T'/T1) in atm p2' = p2·(T'/T2) in atm...............calculate urself.............

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