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Chapter 9: Problem 1252

The pressure and temperature of two different gases \(P\) and T having the volumes \(\mathrm{V}\) for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be, (A) \(\mathrm{P}\) (B) \((\mathrm{P} / 2)\) (C) \(4 \mathrm{P}\) (D) \(2 \mathrm{P}\)

Short Answer

Expert verified
The pressure of the gas mixture when two different gases with the same volume and temperature are mixed, while keeping the same volume and temperature, is \(2P\).

Step by step solution

01

Understand the problem and prerequisites

In this exercise, we are provided with the volume and temperature for two different gases with different pressures. We need to find the pressure of the gas mixture after mixing the two gases, keeping the same volume and the same temperature. To solve this problem, we will use the ideal gas law which is stated as: \(PV = nRT\) Where, P = pressure V = volume n = number of moles R = universal gas constant T = temperature
02

Define the ideal gas law for each individual gas

We are given two different gases, let's denote them as Gas 1 and Gas 2. We can express the ideal gas law for each gas as follows: For Gas 1: \(P_1 V = n_1 RT\) For Gas 2: \(P_2 V = n_2 RT\)
03

Define the gas mixture's properties

When the two gases are mixed, the total volume and temperature remain the same. Therefore, the ideal gas law for the combined gas mixture can be expressed as: \(P_m V = n_m RT\) Where \(P_m\) is the pressure of the mixture and \(n_m\) is the total number of moles in the mixture.
04

Calculate the number of moles in the mixture

Since we are given that the volume and temperature for each individual gas and the gas mixture are the same, we can find the total moles of gases in the mixture by adding the moles in Gas 1 and Gas 2. \(n_m = n_1 + n_2\)
05

Substitute the ideal gas law expressions for the individual gases and the mixture

Now we can substitute the expressions from the ideal gas law for individual gases and the mixture in the equation \(n_m = n_1 + n_2\) \(\frac{P_m V}{RT} = \frac{P_1 V}{RT} + \frac{P_2 V}{RT}\)
06

Simplify the equation and solve for \(P_m\)

Now let's cancel out the common terms on both sides of the equation: \(P_m = P_1 + P_2\) Since both gases have the same pressure P, the pressure of the mixture can be expressed as: \(P_m = P + P\) \(P_m = 2P\) From the given options, we can see that the correct answer is: (D) \(2P\)

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Most popular questions from this chapter

If three molecules have velocities \(0.5,1\) and 2 the ratio of rms speed and average speed is (The velocities are in \(\mathrm{km} / \mathrm{s}\) ) (A) \(0.134\) (B) \(1.34\) (C) \(1.134\) (D) \(13.4\)

At \(\mathrm{O}^{\circ} \mathrm{C}\) the density of a fixed mass of a gas divided by pressure is \(\mathrm{x} .\) At \(100^{\circ} \mathrm{C}\), the ratio will be (A) \(\mathrm{x}\) (B) \([(273) /(373)] \mathrm{x}\) (C) \([(373) /(273)] \mathrm{x}\) (D) \([(100) /(273)] \mathrm{x}\)

When the temperature of a gas is raised from \(27^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the percentage increase in the rms velocity of the molecules will be (A) \(15 \%\) (B) \(17.5 \%\) (C) \(10 \%\) (D) \(20 \%\)

1 mole of gas occupies a volume of \(100 \mathrm{~m} 1\) at \(50 \mathrm{~mm}\) pressure. What is the volume occupied by two moles of gas at $100 \mathrm{~mm}$ pressure and at same temperature (A) \(50 \mathrm{ml}\) (B) \(200 \mathrm{ml}\) (C) \(100 \mathrm{ml}\) (D) \(500 \mathrm{~m} 1\)

The rms speed of gas molecules is given by (A) \(2.5 \sqrt{\left[M_{0} /(R T)\right]}\) (B) \(\left.2.5 \sqrt{[}(\mathrm{RT}) / \mathrm{M}_{0}\right]\) (C) \(\left.1.73 \sqrt{[}(\mathrm{RT}) / \mathrm{M}_{0}\right]\) (D) \(1.73 \sqrt{\left[M_{0} /(R T)\right]}\)
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