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1 mole of a monoatomic gas is mixed with 3 moles of a diatomic gas. What is the molecular specific heat of the mixture at constant volume?
  • a)
    12.5 J / mol K
  • b)
    18.7 J / mol K 
  • c)
    15.2 J / mol K
  • d)
    22.6 J / mol K
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
1 mole of a monoatomic gas is mixed with 3 moles of a diatomic gas. Wh...
Explanation:
for monoatomic gas 
 
from conservation of energy
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Most Upvoted Answer
1 mole of a monoatomic gas is mixed with 3 moles of a diatomic gas. Wh...
Given:
- 1 mole of a monoatomic gas (single atom)
- 3 moles of a diatomic gas (two atoms)

To find:
The molecular specific heat of the mixture at constant volume.

Solution:

1. Concept:
The specific heat capacity of a gas is the amount of heat required to raise the temperature of 1 mole of the gas by 1 Kelvin.

2. Calculation:
To calculate the molecular specific heat of the mixture, we need to consider the specific heat capacity of each gas and their respective mole ratios.

a) Monoatomic gas:
The specific heat capacity of a monoatomic gas at constant volume (Cv) is given by the equation:

Cv = (3/2)R

Where R is the ideal gas constant.

b) Diatomic gas:
The specific heat capacity of a diatomic gas at constant volume (Cv) is given by the equation:

Cv = (5/2)R

c) Mixture of gases:
The total number of moles in the mixture is:
1 mole (monoatomic gas) + 3 moles (diatomic gas) = 4 moles

The specific heat capacity of the mixture (Cv_mixture) is given by the equation:

Cv_mixture = (Cv_monoatomic * n_monoatomic + Cv_diatomic * n_diatomic) / (n_monoatomic + n_diatomic)

Where:
Cv_monoatomic = specific heat capacity of monoatomic gas
n_monoatomic = number of moles of monoatomic gas
Cv_diatomic = specific heat capacity of diatomic gas
n_diatomic = number of moles of diatomic gas

Substituting the values:
Cv_mixture = ((3/2)R * 1 + (5/2)R * 3) / (1 + 3)
Cv_mixture = (3R + 15R) / 4
Cv_mixture = 18R / 4
Cv_mixture = 4.5R

3. Conclusion:
The molecular specific heat of the mixture at constant volume is 4.5 times the ideal gas constant (R). Since the value of R is approximately 8.314 J/mol·K, the molecular specific heat of the mixture is approximately 4.5 * 8.314 J/mol·K = 37.41 J/mol·K.

Answer:
The molecular specific heat of the mixture at constant volume is approximately 37.41 J/mol·K. Therefore, option B (18.7 J/mol·K) is the correct answer.
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1 mole of a monoatomic gas is mixed with 3 moles of a diatomic gas. What is the molecular specific heat of the mixture at constant volume?a)12.5J / mol Kb)18.7 J / mol Kc)15.2J / mol Kd)22.6 J / mol KCorrect answer is option 'B'. Can you explain this answer?
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