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

(II) What is the pressure inside a 38.0-L container holding 105.0 kg of argon gas at 21.6°C?

Short Answer

Expert verified

The pressure onthe inside of the container is \(1.69 \times {10^8}\;{\rm{Pa}}\).

Step by step solution

01

Concepts 

The ideal gas law is \(PV = nRT\) for n mole of ideal gas.

For this problem, first, find the given data in the SI unit and then put it in the ideal gas equation.

02

Given data 

The initial temperature is \(T = {21.6^ \circ }{\rm{C}} = 294.6\;{\rm{K}}\).

The volume of the container is \(V = 38.0\;{\rm{L}} = 38.0 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{3}}}\)

The mass of argon inside the container is \(M = 105.0\;{\rm{kg}}\).

The mass of onemole of argon gas is \(m = 39.95\;{\rm{g/mol}} = 39.95 \times {10^{ - 3}}\;{\rm{kg/mol}}\).

Letbe the pressure of the gas inside the container.

03

Calculation

Now the number of moles of the argon gas is \(n = \frac{M}{m}\).

Then from the ideal gas equation,you get the following:

\(\begin{array}{c}PV = nRT\\PV = \frac{M}{m}RT\\P = \frac{{MRT}}{{mV}}\end{array}\)

Now, substituting the values into the above equation,you getthe following:

\(\begin{array}{c}P = \frac{{\left( {105.0\;{\rm{kg}}} \right) \times \left( {8.314\;{\rm{J/mol}} \cdot {\rm{K}}} \right) \times \left( {294.6\;{\rm{K}}} \right)}}{{\left( {39.95 \times {{10}^{ - 3}}\;{\rm{kg/mol}}} \right) \times \left( {38.0 \times {{10}^{ - 3}}\;{{\rm{m}}^{\rm{3}}}} \right)}}\\ = 1.69 \times {10^8}\;{\rm{Pa}}\end{array}\)

Hence, the pressure onthe inside of the container is \(1.69 \times {10^8}\;{\rm{Pa}}\).

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

Question 17: Escape velocity from the Earth refers to the minimum speed an object must have to leave the Earth and never return. (a) The escape velocity from the Moon is about one-fifth what it is for the Earth, due to the Moon’s smaller mass. Explain why the Moon has practically no atmosphere. (b) If hydrogen was once in the Earth’s atmosphere, why would it have probably escaped?

The escape speed from the Earth is \({\bf{1}}{\bf{.12 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{m/s}}\),that is, a gas molecule traveling away from Earth near the outer boundary of the Earth’s atmosphere would, at this speed, be able to escape from the Earth’s gravitational field and be lost in the atmosphere. At what temperature is the RMS speed of (a) oxygen molecules and (b) helium atoms equal to \({\bf{1}}{\bf{.12 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{m/s}}\)? (c) Can you explain why our atmosphere contains oxygen but not helium?

A student’s weight displayed on a digital scale is \({\bf{117}}{\bf{.2}}\;{\bf{lb}}\). This would suggest her weight is

(a) within\({\bf{1\% }}\)of\({\bf{117}}{\bf{.2}}\;{\bf{lb}}\).

(b) exactly \({\bf{117}}{\bf{.2}}\;{\bf{lb}}\).

(c) somewhere between \({\bf{117}}{\bf{.18}}\;{\bf{lb}}\) and \({\bf{117}}{\bf{.2}}\;{\bf{lb}}\).

(d) somewhere between \({\bf{117}}{\bf{.0}}\) and \({\bf{117}}{\bf{.4}}\;{\bf{lb}}\).

Question 10: The units for the coefficient of linear expansion \({\bf{\alpha }}\) are \({\left( {{\bf{C^\circ }}} \right)^{{\bf{ - 1}}}}\)and there is no mention of a length unit such as meters. Would the expansion coefficient change if we used feet or millimeters instead of meters? Explain.

Question: (II)What is the rms speed of nitrogen molecules contained in an\({\bf{8}}{\bf{.5}}\;{{\bf{m}}^{\bf{3}}}\)volume at 2.9 atm if the total amount of nitrogen is 2100 mol?

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