Question

Internal energy of $n_1$ moles of hydrogen at temperature $T$ is equal to internal energy of $m_2$ moles of helium at temperature $2T$. The ratio $ \frac{n_1}{n_2} $ is

• A. \( \frac{6}{5} \)
• B. \( \frac{3}{7} \)
• C. \( \frac{5}{3} \)
• D. \( \frac{3}{2} \)

Hint:

For gases, the internal energy is directly proportional to the number of moles and the temperature. When comparing the internal energies of different gases, ensure that you account for temperature differences and the type of gas involved.

Answer 1

The Correct Option is A

The internal energy of an ideal gas can be written as: \[ U = \frac{3}{2} n R T \] Where: - \( U \) is the internal energy, - \( n \) is the number of moles, - \( R \) is the gas constant, - \( T \) is the temperature. For hydrogen, the internal energy \( U_1 \) is: \[ U_1 = \frac{3}{2} n_1 R T \] For helium, the internal energy \( U_2 \) is: \[ U_2 = \frac{3}{2} m_2 R (2T) \] Now, we are given that the internal energy of hydrogen is equal to the internal energy of helium: \[ \frac{3}{2} n_1 R T = \frac{3}{2} m_2 R (2T) \] Simplifying the equation: \[ n_1 T = 2 m_2 T \] \[ n_1 = 2 m_2 \] Therefore, the ratio \( \frac{n_1}{n_2} \) is: \[ \frac{n_1}{n_2} = \frac{6}{5} \]
Thus, the ratio \( \frac{n_1}{n_2} \) is \( \frac{6}{5} \).