C&D Problem 20.34 - Enhanced - with Feedback 2.4 mol of monatomic gas A initiall

Question

C&D

Problem 20.34 - Enhanced - with Feedback 2.4 mol of monatomic gas A initially has 4800 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially has 8200 J of thermal energy. You may want to review (LCD pages 559 - 561) Part B What is the final thermal energy of the gas A? Express your answer to two significant figures and include the appropriate units. PuÅ 5 - a ? Es = | Value Units Submit My Answers Give Up Part C What is the final thermal energy of the gas B? Express your answer to two significant figures and include the appropriate units. CHA S o a ? Ez = | Value Units Submit My Answers Give Up

Explanation / Answer

The internal energy of an ideal gas is given by:
E = nCvT
or in terms of molar energy:
e = E/n = CvT

For a monatomic ideal gas Cv = (3/2)R, i.e. it is the same for the two gases. Therefore the gas samples have the same molar energy in equilibrium. Let's call this value e'

Use this to balance initial energy and energy in equilibrium:
E1 + E2 = (n1 + n2)e'
=>
e' = (E1 + E2) / (n1 + n2)
= (4800J + 8200J) / (2.4mol + 3mol) = 2407.407 J/mol

So the amount of energy transferred between the gases is
E1 = n1e' - E1 = 2.4mol2407.407J/mol - 4800J = 977.78 J

So 977.78 J are transferred from gas B to gas A

Final thermal energy of Gas A = 5777.78 J

Final thermal energy of Gas B = 8200 - 977.778 = 7222.22 J

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