Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 33

Complete the following table for an ideal gas: $$\begin{array}{cccc}{P} & {v} & {n} & {T} \\ {2.00 \text { atm }} & {1.00 \mathrm{L}} & {0.500 \mathrm{mol}} & {\text { ?K }} \\ {0.300 \mathrm{atm}} & {0.250 \mathrm{L}} & {? \mathrm{mol}}\\\\{650 \text { torr }} & {\text { ?L }} & {0.333 \mathrm{mol}} & {350 \mathrm{K}} \\ {\text { ? atm }} & {585 \mathrm{mL}} & {0.250 \mathrm{mol}} & {295 \mathrm{K}}\end{array}$$

Short Answer

Expert verified
\( T_1 = 49.4 \mathrm{K}, \: n_2 \: \text{cannot be determined}, \: V_3 = 6.307\: \mathrm{L}, \: P_4 = 3.40 \: \mathrm{atm} \)

Step by step solution

01

Calculate the temperature in the first row

Apply the ideal gas law for the first row: \( P_1V_1 = n_1RT_1 \). Then solve for \(T_1\): \( T_1 = \frac{P_1V_1}{n_1R} \). Now plug in the given values: \( T_1 = \frac{2.00 \mathrm{atm} \cdot 1.00\mathrm{L}}{0.500\mathrm{mol} \cdot 0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}}} \). Step 2: Calculate the moles in the second row
02

Calculate the moles in the second row

Apply the ideal gas law for the second row: \(P_2V_2 = n_2RT_2 \). Then solve for \(n_2\): \( n_2 = \frac{P_2V_2}{RT_2} \). Now plug the given values: \( n_2 = \frac{0.300 \mathrm{atm} \cdot 0.250\mathrm{L}}{0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}} \cdot ? \mathrm{K}} \). Note that we cannot solve for \(n_2\) because the temperature is missing. Step 3: Calculate the volume in the third row
03

Calculate the volume in the third row

Convert the pressure to atmospheres: \( P_3 = \frac{650 \mathrm{torr}}{760 \frac{\mathrm{torr}}{\mathrm{atm}}} \). Apply the ideal gas law for the third row: \(P_3V_3 = n_3RT_3 \). Then solve for \(V_3\): \( V_3 = \frac{n_3RT_3}{P_3} \). Now plug in the values: \( V_3 = \frac{0.333\mathrm{mol} \cdot 0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}} \cdot 350\mathrm{K}}{\frac{650 \mathrm{torr}}{760 \frac{\mathrm{torr}}{\mathrm{atm}}}} \). Step 4: Calculate the pressure in the fourth row
04

Calculate the pressure in the fourth row

Convert the volume to liters: \(V_4 = \frac{585 \mathrm{mL}}{1000 \frac{\mathrm{mL}}{\mathrm{L}}} \). Apply the ideal gas law for the fourth row: \(P_4V_4 = n_4RT_4 \). Then solve for \(P_4\): \( P_4 = \frac{n_4RT_4}{V_4} \). Now plug in the values: \( P_4 = \frac{0.250\mathrm{mol} \cdot 0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}} \cdot 295\mathrm{K}}{\frac{585 \mathrm{mL}}{1000 \frac{\mathrm{mL}}{\mathrm{L}}}} \).

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.

Start your free trial

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) If the pressure exerted by ozone, \(\mathrm{O}_{3},\) in the stratosphere is \(3.0 \times 10^{-3}\) atm and the temperature is 250 \(\mathrm{K}\) , how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately 0.04\(\%\) of Earth's atmosphere. If you collect a \(2.0-\mathrm{L}\) sample from the atmosphere at sea level \((1.00 \mathrm{atm})\) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\text { mfp }}\) , like the ideal-gas constant) and define units for \(R_{\operatorname{mfp}}\) .

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-} .\) (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ If you allow 15.0 \(\mathrm{g}\) of \(\mathrm{NaClO}_{2}\) to react with 2.00 \(\mathrm{L}\) of chlorine gas at a pressure of 1.50 atm at \(21^{\circ} \mathrm{C},\) how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

The physical fitness of athletes is measured by \(^{u} V_{\mathrm{O}_{2}} \max _{2}^{\prime \prime}\) which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of 45 \(\mathrm{mL} \mathrm{O}_{2 / \mathrm{kg}}\) body mass/min, but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of 88.0 \(\mathrm{mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min. (a) Calculate the volume of oxygen, in mL, consumed in 1 by an average man who weighs 185 lbs and has a \(V_{\mathrm{O}_{2}}\) max reading of 47.5 \(\mathrm{mLO}_{2} / \mathrm{kg}\) body mass/min. (b) If this man lost \(20 \mathrm{lb},\) exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to 65.0 \(\mathrm{mL}\) O \(_{2} / \mathrm{kg}\) body mass/min, how many mL of oxygen would he consume in 1 \(\mathrm{hr}\) ?

Nitrogen and hydrogen gases react to form ammonia gas as follows: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ At a certain temperature and pressure, 1.2 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) reacts with 3.6 \(\mathrm{Lof} \mathrm{H}_{2} .\) If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3},\) at the same temperature and pressure, will be produced?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation
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