A container of volume \(200\text{ cm}^3\) contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature 200 K (\(R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}\) ) will be
\(4.15 \times 10^5\text{ Pa}\)
\(4.15 \times 10^6\text{ Pa}\)
\(6.15 \times 10^5\text{ Pa}\)
\(6.15 \times 10^4\text{ Pa}\)
Solution:
Using the ideal gas equation \(P = \frac{nRT}{V}\), with total moles \(n = 0.2 + 0.3 = 0.5\text{ mol}\) and volume \(V = 200 \times 10^{-6}\text{ m}^3\). Calculating gives \(P = \frac{0.5 \times 8.3 \times 200}{2 \times 10^{-4}} = 4.15 \times 10^6\text{ Pa}\).
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