A vessel contains 1 mole of \(\mathrm{O}_{2}\) gas (molar mass 32 ) at a temperature \(\mathrm{T}\). The pressure of the gas is \(\mathrm{P}\). An identical vessel containing one mole of He gas (molar mass 4) at temperature \(2 \mathrm{~T}\) has a pressure of (A) \(2 \mathrm{P}\) (B) \(\mathrm{P}\) (C) \(8 \mathrm{P}\) (D) \((\mathrm{P} / 8)\)

For this exercise, we will use the ideal gas law, which is given as: \[PV = nRT\] For the Oxygen gas in the first vessel: \( P = \) Pressure of the \(\mathrm{O}_{2}\) gas \( V = \) Volume of the \(\mathrm{O}_{2}\) gas \( n = \) Number of moles of the gas (\(1\) mole for \(\mathrm{O}_{2}\)) \( R = \) Ideal gas constant \( T = \) Temperature of the \(\mathrm{O}_{2}\) gas For the Helium gas in the second vessel: \( P_{He} = \) Pressure of the He gas (we need to find this) \( V_{He} = \) Volume of the He gas \( n = \) Number of moles of the gas (\(1\) mole for He) \( R = \) Ideal gas constant \( T_{He} = 2T \) Temperature of the He gas (given as twice the temperature of the \(\mathrm{O}_{2}\) gas)

We can use the ideal gas law to set up a ratio for the pressure between both vessels. Let's rewrite the ideal gas law for each vessel: For \(\mathrm{O}_{2}\) gas: \(PV = nRT \Rightarrow P = \frac{nRT}{V}\) For He gas: \(P_{He}V_{He} = nRT_{He} \Rightarrow P_{He} = \frac{nRT_{He}}{V_{He}}\) Now, we can set up a ratio: \(\frac{P_{He}}{P} = \frac{\frac{nRT_{He}}{V_{He}}}{\frac{nRT}{V}}\)

Now, let's simplify the ratio and find the pressure of the He gas in the second vessel: \[\frac{P_{He}}{P} = \frac{nRT_{He}V}{nRTV_{He}}\] Notice that the volumes of the two vessels are identical, meaning that \(V = V_{He}\). Also, the number of moles is the same for both gases, which is \(1\) mole. We can cancel out these terms in the ratio: \[\frac{P_{He}}{P} = \frac{RT_{He}}{RT}\] Now we can substitute the given information for the gas temperatures: \(T_{He} = 2T\) \[\frac{P_{He}}{P} = \frac{R(2T)}{RT}\] Simplifying the ratio, we find: \[\frac{P_{He}}{P} = 2\] Thus, the pressure of the He gas in the second vessel is twice the pressure of the \(\mathrm{O}_{2}\) gas in the first vessel. Therefore, the correct answer is: (A) \(2 \mathrm{P}\)