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Chapter 19: Problem 6

Two identical containers hold equal masses of gas, oxygen in one and nitrogen in the other. The gases are held at the same temperature. How does the pressure of the oxygen compare to that of the nitrogen? a) \(p_{\mathrm{O}}>p_{\mathrm{N}}\) b) \(p_{\mathrm{O}}=p_{\mathrm{N}}\) c) \(p_{\mathrm{O}}

Short Answer

Expert verified
a) \(p_{\mathrm{O}} > p_{\mathrm{N}}\) b) \(p_{\mathrm{O}} = p_{\mathrm{N}}\) c) \(p_{\mathrm{O}} < p_{\mathrm{N}}\) d) None of the above Answer: c) \(p_{\mathrm{O}} < p_{\mathrm{N}}\)

Step by step solution

01

Write down the Ideal Gas Law equation

The Ideal Gas Law equation is given by: \(PV = nRT\) Where: - P is the pressure of the gas - V is the volume of the container - n is the number of moles of gas - R is the universal gas constant - T is the temperature of the gas
02

Identify the important variables

Since we are comparing the pressure of the gases, we need to find the ratio of the number of moles for each gas: \(n_{\mathrm{O}}\) = number of moles of oxygen \(n_{\mathrm{N}}\) = number of moles of nitrogen We know that the mass of the gases is equal, and the volume and temperature are also the same.
03

Calculate the number of moles for each gas

We can calculate the number of moles for each gas using the following formula: \(n = \frac{m}{M}\) Where: - n is the number of moles - m is the mass of the gas - M is the molar mass of the gas For oxygen: - molar mass M = 32 g/mol For nitrogen: - molar mass M = 28 g/mol Given that the masses of the gases are equal (m), we can write the ratio of the number of moles for each gas as follows: \(\frac{n_\mathrm{O}}{n_\mathrm{N}} = \frac{\frac{m}{32}}{\frac{m}{28}} = \frac{28}{32}\)
04

Use the Ideal Gas Law to compare pressures

We can now use the Ideal Gas Law to compare the pressures of oxygen and nitrogen under the same conditions: \(p_\mathrm{O}V = n_\mathrm{O}RT\) \(p_\mathrm{N}V = n_\mathrm{N}RT\) Since the volumes and temperatures are equal, we can divide both equations: \(\frac{p_\mathrm{O}}{p_\mathrm{N}} = \frac{n_\mathrm{O}}{n_\mathrm{N}} = \frac{28}{32}\)
05

Determine the relationship between the pressures

From the previous step, we can see that the ratio of the pressures is less than 1: \(\frac{p_\mathrm{O}}{p_\mathrm{N}} = \frac{28}{32} < 1\) Which means: \(p_\mathrm{O} < p_\mathrm{N}\) The correct answer is c) \(p_{\mathrm{O}}<p_{\mathrm{N}}\).

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Most popular questions from this chapter

Compare the average kinetic energy at room temperature of a nitrogen molecule to that of a nitrogen atom. Which has the larger kinetic energy? a) nitrogen atom b) nitrogen molecule c) They have the same energy. d) It depends upon the pressure.

Consider a box filled with an ideal gas. The box undergoes a sudden free expansion from \(V_{1}\) to \(V_{2}\). Which of the following correctly describes this process? a) Work done by the gas during the expansion is equal to \(n R T \ln \left(V_{2} / V_{1}\right)\) b) Heat is added to the box. c) Final temperature equals initial temperature times \(\left(V_{2} / V_{1}\right)\). d) The internal energy of the gas remains constant.

An ideal gas may expand from an initial pressure, \(p_{\mathrm{i}},\) and volume, \(V_{\mathrm{i}},\) to a final volume, \(V_{\mathrm{f}}\), isothermally, adiabatically, or isobarically. For which type of process is the heat that is added to the gas the largest? (Assume that \(p_{i}, V_{i}\) and \(V_{f}\) are the same for each process.) a) isothermal process b) adiabatic process c) isobaric process d) All the processes have the same heat flow.

6.00 liters of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergo the following processes: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the pressure, volume, and temperature of the gas in states 2 and \(3 .\) How many moles of the gas are there?

A sealed container contains 1.00 mole of neon gas at STP. Estimate the number of neon atoms having speeds in the range from \(200.00 \mathrm{~m} / \mathrm{s}\) to \(202.00 \mathrm{~m} / \mathrm{s}\). (Hint: Assume the probability of neon atoms having speeds between \(200.00 \mathrm{~m} / \mathrm{s}\) and \(202.00 \mathrm{~m} / \mathrm{s}\) is constant.
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