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Chapter 9: Problem 1224

\(\mathrm{O}_{2}\) gas is filled in a vessel. If pressure is double, temperature becomes four times, how many times its density will become. (A) 4 (B) \((1 / 4)\) (C) 2 (D) \((1 / 2)\)

Short Answer

Expert verified
The density of the gas will become 2 times its initial value. The correct answer is (C) 2.

Step by step solution

01

Write down the ideal gas law and density formulas

The ideal gas law states that \[PV = nRT\] Where: P = pressure V = volume n = the number of moles of gas R = the universal gas constant T = temperature (in Kelvin) The formula for the density (ρ) of a gas is: \[\rho = \frac{m}{V}\] Where: ρ = density m = mass of the gas V = volume of the gas
02

Combine the ideal gas law and density formulas

We will rewrite the number of moles of gas (n) in terms of its mass (m) and molar mass (M) as: \[n = \frac{m}{M}\] Now, substitute this expression for n into the ideal gas law equation: \[PV = \frac{m}{M} RT\] Now we can rewrite the ideal gas law in terms of density (ρ) by taking the mass (m) as the subject: \[m = \frac{M}{R} \cdot \frac{PV}{T}\] Since density is the mass per unit volume, we can divide both sides of the equation by V: \[\rho = \frac{M}{R} \cdot \frac{P}{T}\]
03

Plug in the given changes to the pressure and temperature

We are given that when the pressure (P) is doubled, the temperature (T) becomes four times its original value: \[P_{2} = 2P_{1}\] \[T_{2} = 4T_{1}\] Now, we can calculate the new density (ρ₂) in terms of the initial density (ρ₁): \[\rho_{2} = \frac{M}{R} \cdot \frac{P_{2}}{T_{2}}\] Substitute the given changes to the pressure and temperature: \[\rho_{2} = \frac{M}{R} \cdot \frac{2P_{1}}{4T_{1}}\]
04

Calculate the ratio of the initial density to the new density

First, we can simplify the expression for the new density: \[\rho_{2} = \frac{1}{2} \cdot \frac{M}{R} \cdot \frac{P_{1}}{T_{1}}\] Now, divide the expression for the initial density (ρ₁) by the expression for the new density (ρ₂) to get the ratio: \[\frac{\rho_{1}}{\rho_{2}} = \frac{\frac{M}{R} \cdot \frac{P_{1}}{T_{1}}}{\frac{1}{2} \cdot \frac{M}{R} \cdot \frac{P_{1}}{T_{1}}}\]
05

Solve for the ratio of densities

Cancel out the common terms in the numerator and denominator, and solve for the ratio of densities: \[\frac{\rho_{1}}{\rho_{2}} = \frac{\cancel{\frac{M}{R} \cdot \frac{P_{1}}{T_{1}}}}{\frac{1}{2} \cdot \cancel{\frac{M}{R} \cdot \frac{P_{1}}{T_{1}}}} = \frac{1}{\frac{1}{2}} = 2\] So, the density of the gas will become 2 times its initial value. The correct answer is (C) 2.

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Most popular questions from this chapter

\- If electron tube was sealed oft during manuracture at a pressure of $1 \times 10^{-7}\( torr at \)27^{\circ} \mathrm{C}\(. Its volume is \)100 \mathrm{~cm}^{3}$. The number of molecules that remain in the tube is (density of mercury is \(13.6 \mathrm{gcm}^{-3}\) ) \((1\) torr $=133 \mathrm{~Pa}$ ) (A) \(3.9 \times 10^{11}\) (B) \(3 \times 10^{16}\) (C) \(2 \times 10^{14}\) (D) \(7 \times 10^{11}\)

The pressure and temperature of an ideal gas in a closed vessel are $720 \mathrm{kPa}\( and \)40^{\circ} \mathrm{C}\( respectively. If \)(1 / 4)^{\text {th }}$ of the gas is released from the vessel and the temperature of the remaining gas is raised to \(353^{\circ} \mathrm{C}\), final pressure of the gas is (A) \(1440 \mathrm{kPa}\) (B) \(540 \mathrm{kPa}\) (C) \(1080 \mathrm{kPa}\) (D) \(720 \mathrm{kPa}\)

A gas at \(27{ }^{\circ} \mathrm{C}\) temperature and 30 atmospheric pressure $1 \mathrm{~s}$ allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (A) \(273^{\circ} \mathrm{C}\) (B) \(-173^{\circ} \mathrm{C}\) (C) \(173^{\circ} \mathrm{C}\) (D) \(100^{\circ} \mathrm{C}\)

A sample of gas is at \(0^{\circ} \mathrm{C}\). To what temperature it must be raised in order to double the rms speed of molecule. (A) \(270^{\circ} \mathrm{C}\) (B) \(819^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(1090^{\circ} \mathrm{C}\)

To double the volume of a given mass at an ideal gas at $27^{\circ} \mathrm{C}$ keeping the pressure constant one must raise the temperature in degree centigrade (A) \(54^{\circ}\) (B) \(600^{\circ}\) (C) \(327^{\circ}\) (D) \(270^{\circ}\)
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