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Calculate the relative populations of the J = 1 and J = 0 rotational states of HCl at 25°C? [up to 2 decimal place] (Given: BHCl = 10.591 cm–1) [rounded up to three decimal place]
Correct answer is between '2.680,2.720'. Can you explain this answer?
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Calculate the relative populations of the J = 1 and J = 0 rotational s...
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Calculate the relative populations of the J = 1 and J = 0 rotational s...
To calculate the relative populations of the J = 1 and J = 0 rotational states of HCl at 25°C, we can use the Boltzmann distribution equation:
P(J) = (2J + 1) * e^(-E(J)/kT)
where P(J) is the relative population of the J rotational state, J is the rotational quantum number, E(J) is the energy of the J rotational state, k is the Boltzmann constant, and T is the temperature in Kelvin.
First, we need to determine the energy difference between the J = 1 and J = 0 rotational states. The energy of a rotational state can be calculated using the formula:
E(J) = B * J(J+1)
where B is the rotational constant for HCl.
The rotational constant for HCl is approximately 10.59 cm^-1.
For J = 0:
E(0) = B * 0(0 + 1) = 0
For J = 1:
E(1) = B * 1(1 + 1) = 31.77 cm^-1
Next, we need to convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Now, we can calculate the relative populations:
P(0) = (2*0 + 1) * e^(-E(0) / kT)
= 1 * e^(-0 / (1.38 * 10^-23 J/K * 298.15 K))
= 1 * e^0
= 1
P(1) = (2*1 + 1) * e^(-E(1) / kT)
= 3 * e^(-31.77 cm^-1 / (1.38 * 10^-23 J/K * 298.15 K))
= 3 * e^(-1.38 * 10^23 J / (1.38 * 10^-23 J/K * 298.15 K))
= 3 * e^(-1)
≈ 3 * 0.368
≈ 1.104
Therefore, the relative population of the J = 1 rotational state is approximately 1.104 times higher than the relative population of the J = 0 rotational state.
P(J) = (2J + 1) * e^(-E(J)/kT)
where P(J) is the relative population of the J rotational state, J is the rotational quantum number, E(J) is the energy of the J rotational state, k is the Boltzmann constant, and T is the temperature in Kelvin.
First, we need to determine the energy difference between the J = 1 and J = 0 rotational states. The energy of a rotational state can be calculated using the formula:
E(J) = B * J(J+1)
where B is the rotational constant for HCl.
The rotational constant for HCl is approximately 10.59 cm^-1.
For J = 0:
E(0) = B * 0(0 + 1) = 0
For J = 1:
E(1) = B * 1(1 + 1) = 31.77 cm^-1
Next, we need to convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Now, we can calculate the relative populations:
P(0) = (2*0 + 1) * e^(-E(0) / kT)
= 1 * e^(-0 / (1.38 * 10^-23 J/K * 298.15 K))
= 1 * e^0
= 1
P(1) = (2*1 + 1) * e^(-E(1) / kT)
= 3 * e^(-31.77 cm^-1 / (1.38 * 10^-23 J/K * 298.15 K))
= 3 * e^(-1.38 * 10^23 J / (1.38 * 10^-23 J/K * 298.15 K))
= 3 * e^(-1)
≈ 3 * 0.368
≈ 1.104
Therefore, the relative population of the J = 1 rotational state is approximately 1.104 times higher than the relative population of the J = 0 rotational state.
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Calculate the relative populations of the J = 1 and J = 0 rotational states of HCl at 25°C? [up to 2 decimal place] (Given: BHCl = 10.591 cm–1) [rounded up to three decimal place]Correct answer is between '2.680,2.720'. Can you explain this answer?
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