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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

The temperature of a gas at pressure \(\mathrm{P}\) and volume \(\mathrm{V}\) is \(27^{\circ} \mathrm{C}\) Keeping its volume constant if its temperature is raised to \(927^{\circ} \mathrm{C}\), then its pressure will be (A) \(3 \mathrm{P}\) (B) \(2 \mathrm{P}\) (C) \(4 \mathrm{P}\) (D) \(6 \mathrm{P}\)

If \(\mathrm{rms}\) speed of a gas is $\mathrm{v}_{\mathrm{rms}}=1840 \mathrm{~m} / \mathrm{s}\( and its density \)\rho=8.99 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{3}$, the pressure of the gas will be (A) \(1.01 \times 10^{3} \mathrm{Nm}^{-2}\) (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\) (C) \(1.01 \times 10^{7} \mathrm{Nm}^{-2}\) (D) \(1.01 \mathrm{Nm}^{-2}\)

The relation between two specific heats of a gas is (A) $\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=(\mathrm{R} / \mathrm{J})$ (B) \(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{J}\) (C) $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=(\mathrm{R} / \mathrm{J})$ (D) \(\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=\mathrm{J}\)

A diatomic molecule has how many degrees of freedom (For rigid rotator) (A) 4 (B) 3 (C) 6 (D) 5

The volume of a gas at \(20 \mathrm{C}\) is \(200 \mathrm{ml}\). If the temperature is reduced to \(-20^{\circ} \mathrm{C}\) at constant pressure, its volume will be. (A) \(172.6 \mathrm{~m} 1\) (B) \(17.26 \mathrm{ml}\) (C) \(19.27 \mathrm{ml}\) (D) \(192.7 \mathrm{ml}\)
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