Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 48

Determine the work, in kJ, necessary to compress \(2 \mathrm{~kg}\) of air from \(100 \mathrm{kPa}\) to 4000 \(\mathrm{kPa}\) if the temperature is held constant at \(300^{\circ} \mathrm{C}\). a) \(-1210\) c) \(-932\) b) \(-1105\) d) \(-812\)

Short Answer

Expert verified
Answer: The work done is approximately -1105 kJ.

Step by step solution

01

Convert the temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin. The conversion is done by adding 273.15 to the given temperature. \(T(K) = T(^{\circ}C) + 273.15 = 300 + 273.15 = 573.15 \mathrm{K}\)
02

Calculate the number of moles of air

To calculate the number of moles, we use the ideal gas equation. First, we need to calculate the volume of the air at the initial pressure using: \(V_i = \dfrac{n_iRT}{P_i}\) We also need to find the specific gas constant of air, which is \(R = 0.287 \mathrm{~kJ/kg \cdot K}\). Since we are given the mass of the air, we can use this to find the number of moles (\(n_i\)) of the air: \(n_i = \dfrac{m}{M_{air}}\) Here, \(m = 2 \mathrm{~kg}\) and \(M_{air} = 29 \mathrm{~g/mol}\). Now, we can calculate: \(n_i = \dfrac{2 \times 10^3 \mathrm{~g}}{29 \mathrm{~g/mol}} = 68.97 \mathrm{~mol} \)
03

Determine the initial and final volumes

Now we can determine the initial and final volumes of the air using the ideal gas equation: \(V_i = \dfrac{n_iRT}{P_i} = \dfrac{68.97 \cdot 0.287 \cdot 573.15}{100} = 10.23 \mathrm{~m^3} \) \(V_f = \dfrac{n_iRT}{P_f} = \dfrac{68.97 \cdot 0.287 \cdot 573.15}{4000} = 0.2548 \mathrm{~m^3}\)
04

Calculate the work done

Now, we can calculate the work done using the formula for work in an isothermal process: \(W = nRT \ln\left(\dfrac{V_f}{V_i}\right) = 68.97 \cdot 0.287 \cdot 573.15 \cdot \ln\left(\dfrac{0.2548}{10.23}\right) = -1104.97 \mathrm{~kJ}\) Rounding to the nearest whole number, the work done is \(-1105\) kJ. So, the correct answer is \(\boxed{\text{(b) }-1105}\).

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

12,000\( is borrowed now at 12 percent interest. The first payment is \)\$ 4000\( and is made three years from now. The balance of the debt immediately after the payment is a) \)\$ 4000\( c) \)\$ 12,000\( b) \)\$ 8000\( d) \)\$ 12,860$

For the circuit below, with voltage polarities as shown, KVL in equation form is a) \(v_{1}+v_{2}+v_{3}-v_{4}+v_{5}=0\) b) \(-v_{1}+v_{2}+v_{3}-v_{4}+v_{5}=0\) c) \(v_{1}+v_{2}-v_{3}-v_{4}+v_{5}=0\) d) \(-v_{1}-v_{2}-v_{3}+v_{4}+v_{5}=0\)

An alumnus has given Michigan State University \(\$ 10\) million to build and operate a laboratory. Annual operating cost is estimated to be \(\$ 100,000\). The endowment will earn 6 percent interest. Assume an infinite life for the laboratory and determine how much money may be used for its construction. a) \(\$ 5.00 \times 10^{6}\) c) \(\$ 8.72 \times 10^{6}\) b) \(\$ 8.333106\) d) \(\$ 9.90 \times 10^{6}\)

Find an expression for the maximum range of a projectile with initial velocity \(v_{o}\) at angle \(\theta\) with the horizontal. a) \(\frac{v_{o}^{2}}{g} \sin 2 \theta\) c) \(\frac{v_{o}^{2}}{2 g} \sin \theta\) b) \(\frac{v_{o}^{2}}{2 g} \sin ^{2} \theta\) d) \(\frac{v_{o}^{2}}{g} \cos \theta\)

After a factory has been built near a stream, it is learned that the stream occasionally overflows its banks. A hydrologic study indicates that the probability of flooding is about 1 in 8 in any one year. A flood would cause about \(\$ 20,000\) in damage to the factory. A levee can be constructed to prevent flood damage. Its cost will be \(\$ 54,000\) and its useful life is 30 years. Money can be borrowed at 8 percent interest. If the annual equivalent cost of the levee is less than the annual expectation of flood damage, the levee should be built. The annual expectation of flood damage is (1/8) \(\times\) \(20,000=\$ 2,500\). Compute the annual equivalent cost of the levee. a) \(\$ 1,261\) c) \(\$ 4,320\) b) \(\$ 1,800\) d) \(\$ 4,800\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electricity

Read Explanation

Wave Optics

Read Explanation

Rotational Dynamics

Read Explanation

Fluids

Read Explanation

Electricity and Magnetism

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