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Chapter 9: Problem 1257

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) \(1.5 \mathrm{~atm}\)

Short Answer

Expert verified
The pressure on the hydrogen gas is found to be \(2.0 \mathrm{~atm}\), which corresponds to option (C).

Step by step solution

01

1. Determine the molar mass of hydrogen gas

Hydrogen gas is composed of H2 molecules. The molar mass of H is 1, so the molar mass of H2 is 2 g/mol or \(2\times10^{-3} \mathrm{~kg/mol}\).
02

2. Calculate the root mean square speed from the given data

The root mean square speed of hydrogen molecules is given as \(3180 \mathrm{~ms}^{-1}\).
03

3. Calculate the number of gas molecules

Using the equation for root mean square speed, \(v_{rms} = \sqrt{\frac{3RT}{M}}\), we can find the number of moles. Rearranging and solving for the number of moles, we get: \(n = \frac{MV^2}{3RT} \) Here, - M is the molar mass of hydrogen gas found in step 1, which is \(2\times10^{-3} \mathrm{~kg/mol}\). - V is the root mean square speed, which is \(3180 \mathrm{~ms}^{-1}\). - R is the ideal gas constant \(8.314 \mathrm{J/(mol \cdot K)}\). - T is the temperature of the gas in Kelvin, \(273\mathrm{K}\) (0°C is converted to 273 K).
04

4. Calculate the pressure of hydrogen gas

Using the ideal gas equation, \(PV=nRT\), we calculate the pressure on the hydrogen gas. Here, - n is the number of moles calculated in step 3. - R is the ideal gas constant \(8.314 \mathrm{J/(mol \cdot K)}\). - T is the temperature of the gas in Kelvin, \(273\mathrm{K}\). - V is the volume of the gas chamber, which can be calculated by dividing the given mass of the gas (\(8.99 \times 10^{-2} \mathrm{~kg/m^3}\)) by its molar mass (\(2\times10^{-3} \mathrm{~kg/mol}\)).
05

5. Comparing our result with the given options

Now that we have found the pressure on the hydrogen gas, we compare our result to the given options. We need to convert the pressure into atmospheres, so we divide our result by the conversion factor (\(1 \mathrm{~atm}=1.01 \times 10^{5} \mathrm{Nm^{-2}}\)). With all information, we find the answer and compare it to the available options (A, B, C, or D).

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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) \(13.4\)

Air is pumped into an automobile tube upto a pressure of \(200 \mathrm{kPa}\) in the morning when the air temperature is \(22^{\circ} \mathrm{C}\). During the day, temperature rises to \(42^{\circ} \mathrm{C}\) and the tube expands by $2 \%$ The pressure of the air in the tube at this temperature will be approximately. (A) \(209 \mathrm{kPa}\) (B) \(206 \mathrm{kPa}\) (C) \(200 \mathrm{kPa}\) (D) \(212 \mathrm{kPa}\)

A gas at \(27^{\circ} \mathrm{C}\) has a volume \(\mathrm{V}\) and pressure \(\mathrm{P}\). On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be (A) \(1527^{\circ} \mathrm{C}\) (B) \(600^{\circ} \mathrm{C}\) (C) \(162^{\circ} \mathrm{C}\) (D) \(1800^{\circ} \mathrm{C}\)

The temperature at which the rms speed of hydrogen molecules is equal to escape velocity on earth surface will be (A) \(5030 \mathrm{~K}\) (B) \(10063 \mathrm{~K}\) (C) \(1060 \mathrm{~K}\) D) \(8270 \mathrm{~K}\)

The root mean square velocity of the molecules in a sample of helium is $(5 / 7)^{\text {th }}$ that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is \(0^{\circ} \mathrm{C}\), then the temperature of the helium sample is about (A) \(273^{\circ} \mathrm{C}\) (B) \(0^{\circ} \mathrm{C}\) (C) \(0^{\circ} \mathrm{K}\) (D) \(100^{\circ} \mathrm{C}\)
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