2 moles of monoatomic gas at temperature 27degree is mixed with 3 moles of diatomic gas at temperature 127 degree then temperature of mixture is approximately
Question
Question
2 moles of monoatomic gas at temperature 27°C is mixed with 3 moles of diatomic gas at temperature 127°C. Then the temperature of the mixture is approximately?
Solution
To find the approximate temperature of the mixture, we can use the principle of energy conservation.
First, we need to calculate the total energy of each gas before they are mixed. The energy of a monoatomic gas can be calculated using the formula:
E = (3/2) * n * R * T
where E is the total energy, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
For the monoatomic gas: E1 = (3/2) * 2 * R * (27 + 273) = 5R * 300
For the diatomic gas: E2 = (5/2) * 3 * R * (127 + 273) = 15R * 400
Next, we need to calculate the total energy of the mixture. Since energy is conserved, the total energy before and after mixing should be the same.
E_total = E1 + E2
Now, we can find the temperature of the mixture using the formula:
T_mixture = E_total / [(3/2) * (2 + 3) * R]
T_mixture = (5R * 300 + 15R * 400) / (7/2) * R
T_mixture = (1500R + 6000R) / (7/2) * R
T_mixture = 7500R / (7/2) * R
T_mixture = 7500 / (7/2)
T_mixture ≈ 2142.86 degrees
Therefore, the approximate temperature of the mixture is approximately 2142.86 degrees.
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