What should be the minimum volume of the room, if no liquid water is to be present finally in the ro — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Moles of 5 kg water $= 5000 / 18 = 277.78 \text{ mol}$. Volume $V = \frac{nRT}{P} = \frac{277.78 \times 0.08 \times 300}{0.04} = 166668 \text{ L} = 166.67 \text{ m}^3$. Therefore, correct answer is C.