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Chapter 9: Problem 1309

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) \(1.01 \mathrm{Nm}^{-2}\)

Short Answer

Expert verified
The pressure of the gas is calculated using the equation \(P = \rho \cdot \frac{v_{\text{rms}}^2}{3}\), where \(\rho\) is the density and \(v_{\text{rms}}\) is the rms speed. Substituting the given values, we get \(P = (8.99 \times 10^{-2} \mathrm{~kg}\, / \mathrm{m}^{3}) \cdot \frac{(1840\mathrm{~m}\, / \mathrm{s})^2}{3} = 1.01 \times 10^{5} \mathrm{Nm}^{-2}\). Therefore, the correct answer is (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\).

Step by step solution

01

Calculate the number of moles in the gas

To calculate the number of moles in the gas, we need to use the following relationship between rms speed (v_rms), molar mass (M), and density: \[ \rho = \frac{nM}{V}. \] As the density is given, we can calculate the number of moles \(n\) per unit volume \(V\). Rearrange the equation for \(n\): \[ n = \frac{\rho V}{M}. \] We need to find the molar mass (M) next. To find the molar mass, we will use the rms speed and the relationship: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \] where \(R\) is the gas constant and \(T\) is the temperature. Rearrange the equation for molar mass: \[ M = \frac{3RT}{v_{\text{rms}}^2}. \]
02

Calculate the pressure

Now, we will use the ideal gas equation to find the pressure: \[ PV = nRT. \] Plug the expression for \(n\) in terms of \(\rho\), \(V\), and \(M\): \[ P = \frac{nRT}{V} = \frac{\rho R T}{M}. \] We know the value of \(\rho\), \(v_{\text{rms}}\), and will replace \(M\) with the expression from step 1: \[ P = \frac{\rho R T}{\frac{3RT}{v_{\text{rms}}^2}} = \rho \cdot \frac{v_{\text{rms}}^2}{3}. \] Now, we can calculate the pressure using the rms speed and density: \[ P = (8.99 \times 10^{-2} \mathrm{~kg}\, / \mathrm{m}^{3}) \cdot \frac{(1840\mathrm{~m}\, / \mathrm{s})^2}{3} = 1.01 \times 10^{5} \mathrm{Nm}^{-2}. \] So, the correct answer is (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\).

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Most popular questions from this chapter

Suppose ideal gas equation follows \(V P^{3}=\) constant, Initial temperature and volume of the gas are \(\mathrm{T}\) and \(\mathrm{V}\) respectively. If gas expand to \(27 \mathrm{~V}\), then temperature will become (A) \(9 \mathrm{~T}\) (B) \(27 \mathrm{~T}\) (C) (T/9) (D) \(\mathrm{T}\)

The root mean square speed of hydrogen molecules of an ideal hydrogen kept in a gas chamber at \(0^{\circ} \mathrm{C}\) is \(3180 \mathrm{~ms}^{-1}\). The pressure on the hydrogen gas is (Density of hydrogen gas is $8.99 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{3}, 1 \mathrm{~atm}=1.01 \times 10^{5} \mathrm{Nm}^{-2}$ ) (A) \(1.0 \mathrm{~atm}\) (B) \(3.0 \mathrm{~atm}\) (C) \(2.0 \mathrm{~atm}\) (D) \(1.5 \mathrm{~atm}\)

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) \(720 \mathrm{kPa}\)

For a gas, the rms speed at \(800 \mathrm{~K}\) is (A) Four times the value at \(200 \mathrm{~K}\) (B) Twice the value at \(200 \mathrm{~K}\) (C) Half the value at \(200 \mathrm{~K}\) (D) same as at \(200 \mathrm{~K}\)

Under constant temperature, graph between \(\mathrm{p}\) and \((1 / \mathrm{V})\) is (A) Hyperbola (B) Circle (C) Parabola (D) Straight line
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