Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which of the following gases has the highest root-mean-square speed? a) nitrogen at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) b) argon at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) c) argon at 2 atm and \(30^{\circ} \mathrm{C}\) d) oxygen at 2 atm and \(30^{\circ} \mathrm{C}\) e) nitrogen at 2 atm and \(15^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: Nitrogen at 1 atm and 30°C has the highest root-mean-square speed, with a value of 516.98 m/s.

Step by step solution

01

Convert temperatures to Kelvin

We'll convert the given temperatures in Celsius to Kelvin by adding 273.15 to each temperature. For the gases at \(30^{\circ} \mathrm{C}\): \(T_1 = 30 + 273.15 = 303.15 \mathrm{K}\) For the gases at \(15^{\circ} \mathrm{C}\): \(T_2 = 15 + 273.15 = 288.15 \mathrm{K}\)
02

Determine molar masses

We need to find the molar mass of each gas in kg/mol. a) Nitrogen: (N2) \(M_a = 28 \times 10^{-3} \mathrm{kg/mol}\) b) Argon: (Ar) \(M_b = 40 \times 10^{-3} \mathrm{kg/mol}\) d) Oxygen: (O2) \(M_d = 32 \times 10^{-3} \mathrm{kg/mol}\) (Nitrogen and Oxygen are diatomic molecules.)
03

Calculate RMS speed

Now, we'll use the equation \(V_{rms} = \sqrt{\frac{3RT}{M}}\) to calculate RMS speed for each gas. Note that for this problem, the pressure doesn't affect RMS speed, which can be neglected. - The gas constant \(R = 8.314 \mathrm{J/(mol \cdot K)}\) \(V_{rms-a} = \sqrt{\frac{3 \times 8.314 \times 303.15}{28\times10^{-3}}}\) \(V_{rms-b} = \sqrt{\frac{3 \times 8.314 \times 303.15}{40\times10^{-3}}}\) \(V_{rms-c} = \sqrt{\frac{3 \times 8.314 \times 303.15}{40\times10^{-3}}}\) \(V_{rms-d} = \sqrt{\frac{3 \times 8.314 \times 303.15}{32\times10^{-3}}}\) \(V_{rms-e} = \sqrt{\frac{3 \times 8.314 \times 288.15}{28\times10^{-3}}}\)
04

Compare RMS speeds

We can now compare the calculated RMS speeds to find the highest value: \(V_{rms-a} = 516.98 \mathrm{m/s}\) \(V_{rms-b} = 431.76 \mathrm{m/s}\) \(V_{rms-c} = 431.76 \mathrm{m/s}\) \(V_{rms-d} = 483.58 \mathrm{m/s}\) \(V_{rms-e} = 499.69 \mathrm{m/s}\) Clearly, nitrogen at \(1\mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) (option a) has the highest RMS speed of 516.98 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A relationship that gives the pressure, \(p\), of a substance as a function of its density, \(\rho\), and temperature, \(T\), is called an equation of state. For a gas with molar mass \(M\), write the Ideal Gas Law as an equation of state.

One hundred milliliters of liquid nitrogen with a mass of \(80.7 \mathrm{~g}\) is sealed inside a 2 - \(L\) container. After the liquid nitrogen heats up and turns into a gas, what is the pressure inside the container? a) 0.05 atm d) 9.1 atm b) 0.08 atm e) 18 atm c) \(0.09 \mathrm{~atm}\)

A polyatomic ideal gas occupies an initial volume \(V_{\mathrm{i}}\), which is decreased to \(\frac{1}{8} V_{\mathrm{i}}\) via an adiabatic process. Which equation expresses the relationship between the initial and final temperatures of this gas? a) \(T_{\mathrm{f}}=\frac{2}{3} T_{\mathrm{i}}\) d) \(T_{\mathrm{f}}=\frac{9}{5} T_{\mathrm{i}}\) b) \(T_{\mathrm{f}}=2 T_{\mathrm{i}}\) e) \(T_{\mathrm{f}}=8 T_{\mathrm{i}}\) c) \(T_{\mathrm{f}}=16 T_{\mathrm{i}}\)

A 3.787 - \(L\) bottle contains air plus \(n\) moles of sodium bicarbonate and \(n\) moles of acetic acid. These compounds react to produce \(n\) moles of carbon dioxide gas, along with water and sodium acetate. The bottle is tightly sealed at atmospheric pressure \(\left(1.013 \cdot 10^{5} \mathrm{~Pa}\right)\) before the reaction occurs. The pressure inside the bottle when the reaction is complete is \(9.599 \cdot 10^{5} \mathrm{~Pa}\). How many moles, \(n\), of carbon dioxide gas are in the bottle? Assume that the bottle is kept in a water bath that keeps the temperature in the bottle constant.

At a temperature of \(295 .\) K, the vapor pressure of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) is \(60.7 \mathrm{kPa}\). Suppose \(1.000 \mathrm{~g}\) of gaseous pentane is contained in a cylinder with diathermal (thermally conducting) walls and a piston to vary the volume. The initial volume is \(1.000 \mathrm{~L}\), and the piston is moved in slowly, keeping the temperature at \(295 \mathrm{~K}\). At what volume will the first drop of liquid pentane appear?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free