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Three moles of nitrogen is mixed with six moles of helium. The effective molar specific heat of the mixture at constant volume is
- a)1.7 R
- b)1.4 R
- c)1.83 R
- d)1.65 R
Correct answer is option 'C'. Can you explain this answer?
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Three moles of nitrogen is mixed with six moles of helium. The effect...
Explanation:
Molar specific heat at constant volume (Cv) for a gas is given by the formula
Cv = (f/2)R
Where f is the degree of freedom of the gas molecule and R is the gas constant.
For monoatomic gases like helium, f = 3 and for diatomic gases like nitrogen, f = 5.
To find the effective molar specific heat of the mixture, we need to first calculate the total number of moles and the mole fraction of each gas in the mixture.
Total number of moles = 3 + 6 = 9
Mole fraction of nitrogen = 3/9 = 1/3
Mole fraction of helium = 6/9 = 2/3
The effective molar specific heat of the mixture at constant volume is given by the formula
Cv,mix = ΣxiCi
Where xi is the mole fraction of each gas and Ci is the molar specific heat at constant volume of each gas.
For nitrogen, Cv = (5/2)R and for helium, Cv = (3/2)R.
Substituting the values, we get
Cv,mix = (1/3) x (5/2)R + (2/3) x (3/2)R
Cv,mix = (5/6)R + R
Cv,mix = (11/6)R
Converting R to J/mol-K, we get
Cv,mix = (11/6) x 8.314 J/mol-K
Cv,mix = 15.17 J/mol-K
Finally, we need to convert the answer to the given unit of R. Since R = 8.314 J/mol-K, we get
Cv,mix = 15.17/8.314 R
Cv,mix = 1.83 R
Therefore, the correct answer is option (C) 1.83 R.
Molar specific heat at constant volume (Cv) for a gas is given by the formula
Cv = (f/2)R
Where f is the degree of freedom of the gas molecule and R is the gas constant.
For monoatomic gases like helium, f = 3 and for diatomic gases like nitrogen, f = 5.
To find the effective molar specific heat of the mixture, we need to first calculate the total number of moles and the mole fraction of each gas in the mixture.
Total number of moles = 3 + 6 = 9
Mole fraction of nitrogen = 3/9 = 1/3
Mole fraction of helium = 6/9 = 2/3
The effective molar specific heat of the mixture at constant volume is given by the formula
Cv,mix = ΣxiCi
Where xi is the mole fraction of each gas and Ci is the molar specific heat at constant volume of each gas.
For nitrogen, Cv = (5/2)R and for helium, Cv = (3/2)R.
Substituting the values, we get
Cv,mix = (1/3) x (5/2)R + (2/3) x (3/2)R
Cv,mix = (5/6)R + R
Cv,mix = (11/6)R
Converting R to J/mol-K, we get
Cv,mix = (11/6) x 8.314 J/mol-K
Cv,mix = 15.17 J/mol-K
Finally, we need to convert the answer to the given unit of R. Since R = 8.314 J/mol-K, we get
Cv,mix = 15.17/8.314 R
Cv,mix = 1.83 R
Therefore, the correct answer is option (C) 1.83 R.
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Three moles of nitrogen is mixed with six moles of helium. The effective molar specific heat of the mixture at constant volume isa)1.7 Rb)1.4 Rc)1.83 Rd)1.65 RCorrect answer is option 'C'. Can you explain this answer?
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