/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Suppose you are given two 1-L fl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose you are given two 1-L flasks and told that one contains a gas of molar mass 30 , the other a gas of molar mass 60 , both at the same temperature. The pressure in flask A is \(\mathrm{X}\) atm, and the mass of gas in the flask is \(1.2 \mathrm{~g}\). The pressure Which flask contains the gas of molar mass 30 , and which contains the gas of molar mass 60 ?

Short Answer

Expert verified
Flask A contains the gas with a molar mass of 30 g/mol, and Flask B contains the gas with a molar mass of 60 g/mol.

Step by step solution

01

Recall the Ideal Gas Law

The Ideal Gas Law relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas through the equation: \[ PV = nRT, \] where R is the ideal gas constant, with a value of \( 8.3145 \frac{J}{mol K} \) or \(0.08206 \frac{L atm}{mol K}\). Step 2: Convert the mass to number of moles
02

Convert the mass to the number of moles

We'll need to relate the given mass of the gas (in grams) with a number of moles (n) to eventually use it in the Ideal Gas Law. For this, we'll use the molar mass (M) of the substance: \[ n = \frac{m}{M}, \] where n is the number of moles, m is the mass of the substance, and M is its molar mass. Step 3: Write an equation for each flask
03

Write equations for each flask

Let's consider flask A containing the gas with molar mass 30 and flask B containing a gas with molar mass 60. Given the Pressure, we'll use the Ideal Gas Law to write an equation for each flask: \[ P_A V = \frac{m_A}{30} RT, \] \[ P_B V = \frac{m_B}{60} RT, \] where \(P_A\) is the pressure , \(m_A\) is the mass of gas in flask A, and \(P_B\) and \(m_B\) are flask B pressure and mass. Step 4: Use the given information to find the Pressure ratios
04

Use the given information to find the Pressure ratios

We'll now use the given information about flask A and flask B: flask A's pressure is X atm and mass is 1.2 g. \[ P_A V = \frac{1.2}{30} RT, \] flask B's pressure is 0.5X atm and mass is 1.2 g. \[ P_B V = \frac{1.2}{60} RT, \] Step 5: Solve for the pressure ratios
05

Solve for the pressure ratios

Now we'll compare both equations to find the pressure ratio: \[ \frac{P_B V}{P_A V} = \frac{\frac{1.2}{60} RT}{\frac{1.2}{30} RT}, \] Cancelling similar terms and solving for the pressure ratio, we get: \[ \frac{P_B}{P_A} = \frac{1}{2}, \] Step 6: Determine the flasks' content.
06

Determine the flasks' content.

Since we already know that the pressure ratio between flask A and flask B is equal to 1/2, we can see that: Flask A: Pressure = X atm and Gas with Molar Mass = 30 g/mol Flask B: Pressure = 0.5 X atm and Gas with Molar Mass = 60 g/mol So, flask A contains gas of molar mass 30, and the flask B contains the gas of molar mass 60.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Molar mass is a key concept in chemistry that helps us understand the amount of matter in a given substance. It is defined as the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol).
For example, if you have a substance with a molar mass of 30 g/mol, it means that each mole of this substance weighs 30 grams. Knowing the molar mass allows you to convert between mass and the number of moles using the formula:
  • \( n = \frac{m}{M} \),
where \( n \) is the number of moles, \( m \) is the mass of the substance, and \( M \) is its molar mass.
This conversion is essential when you want to understand how much of a gas you have in a chemical reaction or a container. In our example, flask A with a molar mass of 30 g/mol will have a different number of moles at the same mass compared to flask B with a molar mass of 60 g/mol.
Pressure Ratio
When dealing with gases, the term pressure ratio is crucial, especially in comparative studies of different gases.
The pressure ratio is the comparison of the pressure values between two instances or flasks, measuring how much one pressure is in relation to another.
In the given exercise, by comparing the pressures in flasks A and B, we determine that the ratio \( \frac{P_B}{P_A} = \frac{1}{2} \), meaning that the pressure in flask B is half that in flask A.
  • Flask A has a pressure of \( X \) atm.
  • Flask B has a pressure of \( 0.5X \) atm.
Understanding these pressure ratios helps in identifying specific properties of gases within containers, such as understanding how different molar masses influence the behavior of gases under similar conditions.
Number of Moles
The number of moles is a fundamental measurement in chemistry, signifying the amount of substance present. It represents a specific quantity, Avogadro's number, approximately \( 6.022 \times 10^{23} \) entities (such as atoms or molecules).
In the context of the Ideal Gas Law, the number of moles is crucial because it allows you to relate physical quantities like pressure, volume, and temperature.
  • Using the formula \( n = \frac{m}{M} \), you find the number of moles from mass \( m \) and molar mass \( M \).
In the exercise, converting the mass of the gas in each flask to moles was essential to apply the Ideal Gas Law. This conversion informed us about how much gas was present and allowed for comparison based on different molar masses, leading to the determination of which flask contained which gas.
Gas Properties
Gases have unique properties that distinguish them from liquids and solids, primarily their ability to expand and fill a container, regardless of the container's size.
A few of the key properties of gases include:
  • Pressure: Gases exert pressure uniformly on the walls of their container.
  • Temperature: Typically, the higher the temperature, the more kinetic energy the gas molecules have, influencing pressure and volume.
  • Volume: A gas will expand to occupy the entire volume of its container.
These properties are interconnected through the Ideal Gas Law, \( PV = nRT \), which helps predict and measure gas behavior under various conditions.
In this particular exercise, understanding these properties was crucial, as it provided insight into how we could use the Ideal Gas Law to compare the behavior of gases with different molar masses at different pressures.

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

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\) ? (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL}\); the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

The physical fitness of athletes is measured by \({ }^{~} V_{\mathrm{O}_{2}}\) max, " 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 \(1 \mathrm{hr}\) by an average man who weighs \(185 \mathrm{lbs}\) and has a \(V_{\mathrm{O}_{2}}\) max reading of \(47.5 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min. (b) If this man lost \(20 \mathrm{lb}\), exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to \(65.0 \mathrm{~mL} \mathrm{O} / \mathrm{kg}\) body mass/min, how many \(\mathrm{mL}\) of oxygen would he consume in \(1 \mathrm{hr}\) ?

The typical atmospheric pressure on top of Mt. Everest \((29,028 \mathrm{ft})\) is about 265 torr. Convert this pressure to (a) atm, (b) \(\mathrm{mm} \mathrm{Hg}\), (c) pascals, (d) bars, (e) psi.

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations. Consider a 1000 -megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, \(1.00 \mathrm{~atm}\), and \(27^{\circ} \mathrm{C}\), calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and \(120 \mathrm{~atm}\) and a density of \(1.2 \mathrm{~g} / \mathrm{cm}^{3}\), what volume does it possess? (c) If it is stored underground as a gas at \(36^{\circ} \mathrm{C}\) and \(90 \mathrm{~atm}\), what volume does it occupy?

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$ \mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g) $$ If \(159 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of \(\mathrm{Zn}\) have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

See all solutions

Recommended explanations on Chemistry 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.