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Two boxes A and B having their volume ra...

Two boxes A and B having their volume ratio 1: 6 and filled with Ne are inter connected through a narrow tube a negligble volume Box. A is kept at 200 K and box B at 400K . The ratio of mole of Ne gas in box B to box A is _______________

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To find the ratio of moles of Ne gas in box B to box A, we can use the ideal gas law, which states that \( PV = nRT \). ### Step-by-step Solution: 1. **Identify the Volume Ratio**: Given that the volume ratio of box A to box B is \( V_A : V_B = 1 : 6 \), we can express this as: \[ \frac{V_A}{V_B} = \frac{1}{6} \] 2. **Write the Ideal Gas Law for Both Boxes**: For box A: \[ P_A V_A = n_A R T_A \implies P_A = \frac{n_A R T_A}{V_A} \] For box B: \[ P_B V_B = n_B R T_B \implies P_B = \frac{n_B R T_B}{V_B} \] 3. **Set Up the Ratio of Moles**: We want to find the ratio \( \frac{n_B}{n_A} \). From the ideal gas law, we can express the moles in terms of pressure, volume, and temperature: \[ \frac{n_A}{n_B} = \frac{P_A V_A}{P_B V_B} \cdot \frac{T_B}{T_A} \] 4. **Assume Pressures are Equal**: Since the boxes are interconnected, we can assume that the pressures \( P_A \) and \( P_B \) are equal at equilibrium. Therefore, we can simplify the ratio: \[ \frac{n_A}{n_B} = \frac{V_A}{V_B} \cdot \frac{T_B}{T_A} \] 5. **Substitute Known Values**: We know: - \( V_A = 1 \) - \( V_B = 6 \) - \( T_A = 200 \, K \) - \( T_B = 400 \, K \) Plugging in these values: \[ \frac{n_A}{n_B} = \frac{1}{6} \cdot \frac{400}{200} = \frac{1}{6} \cdot 2 = \frac{1}{3} \] 6. **Find the Ratio \( \frac{n_B}{n_A} \)**: To find \( \frac{n_B}{n_A} \), we take the reciprocal of \( \frac{n_A}{n_B} \): \[ \frac{n_B}{n_A} = 3 \] ### Final Answer: The ratio of moles of Ne gas in box B to box A is: \[ \frac{n_B}{n_A} = 3 : 1 \]
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