Question 48: Thermodynamics & KTG

Question

Internal energy of $$\mathrm{n}_1$$ moles of hydrogen at temperature T is equal to internal energy of $$\mathrm{m}_2$$ moles of helium at temperature 2T. The ratio $$\frac{n_1}{n_2}$$ is

Accepted Answer

To find the ratio $$\frac{n_1}{n_2}$$ of moles of hydrogen to helium, we need to use the formula for the internal energy of an ideal gas. For a gas, internal energy (U) is given by:
$$U = \frac{f}{2} nRT$$
where:
$$f$$ is the degrees of freedom
$$n$$ is the number of moles
$$R$$ is the universal gas constant
$$T$$ is the temperature
For diatomic hydrogen (H
2
), the degrees of freedom at room temperature are 5 (3 translational + 2 rotational), so the internal energy is:
$$U_{H_2} = \frac{5}{2} n_1 R T$$
For monatomic helium (He), the degrees of freedom are 3 (all translational), so the internal energy is:
$$U_{He} = \frac{3}{2} n_2 R \cdot 2T = 3 n_2 R T$$
We are given that the internal energy of $$\mathrm{n}_1$$ moles of hydrogen at temperature $$T$$ is equal to the internal energy of $$\mathrm{n}_2$$ moles of helium at temperature $$2T$$. Therefore, we set the internal energies equal to each other:
$$\frac{5}{2} n_1 R T = 3 n_2 R T$$
We can cancel $$R$$ and $$T$$ from both sides of the equation:
$$\frac{5}{2} n_1 = 3 n_2$$
Solving for the ratio $$\frac{n_1}{n_2}$$, we multiply both sides by 2:
$$5 n_1 = 6 n_2$$
Dividing both sides by 6:
$$\frac{n_1}{n_2} = \frac{6}{5}$$
Therefore, the correct answer is:
Option A: $$\frac{6}{5}$$

Answer

$$\frac{6}{5}$$

Answer

$$\frac{3}{7}$$

Answer

$$\frac{5}{3}$$

Answer

$$\frac{3}{2}$$

Thermodynamics & KTG

Internal energy of $\mathrm{n}_1 moles of hydrogen at temperature T is equal to internal energy of \mathrm{m}_2 moles of helium at temperature 2T. The ratio \frac{n_1}{n_2}$ is

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