Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 117

Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose \(120.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}(g)\) is stored in a \(1100.0-\mathrm{L}\) metal cylinder at \(280^{\circ} \mathrm{C}\). (a) Calculate the pressure of the gas, assuming ideal-gas behavior. (b) By using the data in Table 10.3 , calculate the pressure of the gas according to the van der Waals equation. (c) Under the conditions of this problem, which correction dominates, the one for finite volume of gas molecules or the one for attractive interactions?

Short Answer

Expert verified
The pressure of nitrogen gas using the ideal gas law is calculated as \(P = \frac{nRT}{V}\), after converting the mass of nitrogen gas to moles and temperature to Kelvin. The pressure using the van der Waals equation is calculated as \(P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\), where a and b are the van der Waals constants for nitrogen gas. Comparing the calculated pressures from both equations allows us to determine whether the finite volume term \((n^2a/V^2)\) or the attractive interaction term \((nb)\) correction dominates under these conditions.

Step by step solution

01

(Step 1: Convert mass of nitrogen gas to moles)

To calculate the pressure of nitrogen gas, we first need to convert the given mass (120.00 kg) to moles. We can do this using the molar mass of nitrogen gas, which is 28.02 g/mol. Remember to convert kg to g. Moles of nitrogen gas = \(\frac{120.00 kg \times 1000 \frac{g}{kg}}{28.02 \frac{g}{mol}}\)
02

(Step 2: Convert Celsius to Kelvin)

Pressure calculations require temperature to be expressed in Kelvin. Convert the given temperature from Celsius to Kelvin like this: \(T (K) = T (°C) + 273.15\) \(T (K) = 280 + 273.15\)
03

(Step 3: Calculate pressure using the ideal gas law)

We can now calculate the pressure of nitrogen gas using the ideal gas law: \(PV = nRT\) where: P = pressure (unknown) V = volume = 1100.0 L n = moles of nitrogen gas (calculated in Step 1) R = gas constant = 0.0821 \(\frac{L \cdot atm}{mol \cdot K}\) T = temperature in Kelvin (calculated in Step 2) Solve for pressure (P): \(P = \frac{nRT}{V}\)
04

(Step 4: Calculate pressure using the van der Waals equation)

Now, we'll calculate the pressure using the van der Waals equation instead: \(P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\) where: P = pressure (unknown) n = moles of nitrogen gas (calculated in Step 1) R = gas constant = 0.0821 \(\frac{L \cdot atm}{mol \cdot K}\) T = temperature in Kelvin (calculated in Step 2) V = volume = 1100.0 L a and b = van der Waals constants for nitrogen gas, consult Table 10.3 for their values Solve for pressure (P):
05

(Step 5: Compare the two pressures and determine the dominant correction)

Compare the calculated pressures from Steps 3 and 4 to determine which correction, finite volume or attractive interactions, dominates under these conditions. This can be done by analyzing the differences in pressure values calculated using the ideal gas law and the van der Waals equation. If the difference is largely due to the finite volume term \((n^2a/V^2)\) or the attractive interaction term \((nb)\), the correction that dominates can be determined.

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 6.0-L tank is filled with helium gas at a pressure of 2 MPa. How many balloons (each \(2.00 \mathrm{~L}\) ) can be inflated to a pressure of $101.3 \mathrm{kPa}$, assuming that the temperature remains constant and that the tank cannot be emptied below \(101.3 \mathrm{kPa}\) ?

Gaseous iodine pentafluoride, \(\mathrm{IF}_{5}\), can be prepared by the reaction of solid iodine and gaseous fluorine: $$\mathrm{I}_{2}(s)+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g)$$ A \(5.00-\mathrm{L}\) flask containing \(10.0 \mathrm{~g}\) of \(\mathrm{I}_{2}\) is charged with \(10.0 \mathrm{~g}\) of \(\mathrm{F}_{2},\) and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\). (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask (c) Draw the Lewis structure of IF \(_{5}\). (d) What is the total mass of reactants and products in the flask?

Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$\mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g)$$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate $145 \mathrm{~L}\( of \)\mathrm{H}_{2}\( gas if the pressure of \)\mathrm{H}_{2}$ is \(110 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C} ?\)

Suppose you are given two 2 -L flasks and told that one contains a gas of molar mass 28 , the other a gas of molar mass 56 , both at the same temperature and pressure. The mass of gas in the flask \(A\) is $1.0 \mathrm{~g}\( and the mass of gas in the flask \)\mathrm{B}\( is \)2.0 \mathrm{~g}$. Which flask contains the gas of molar mass 28 , and which contains the gas of molar mass 56 ?

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of $5.67 \mathrm{~L}\( has a partial pressure of \)\mathrm{O}_{2}\( of \)7.066 \mathrm{mPa}\( at \)30^{\circ} \mathrm{C}$, what mass of magnesium will react according to the following equation? $$2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s).$$
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Inorganic 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