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Chapter 8: Problem 1137

\(\mu\) moles of gas expands from volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}\) at constant temperature \(\mathrm{T}\). The work done by the gas is (A) \(\mu \mathrm{RT}\left\\{\mathrm{V}_{2} / \mathrm{V}_{1}\right\\}\) (B) \(\mu \operatorname{RTln}\left\\{\mathrm{V}_{2} / \mathrm{V}_{1}\right\\}\) (C) $\mu \mathrm{RT}\left\\{\left(\mathrm{V}_{\mathrm{v}} / \mathrm{V}_{1}\right)-1\right\\}$ (D) $\mu \operatorname{RTln}\left\\{\left(\mathrm{V}_{2} / \mathrm{V}_{1}\right)\right.$

Short Answer

Expert verified
The short answer is: \(W = nRT \ln{\dfrac{V_2}{V_1}}\), which corresponds to answer choice (B).

Step by step solution

01

Understand the relationship between work, pressure, and volume

When a gas expands, it does work on its surroundings. The work done by a gas when it expands from an initial volume \(V_1\) to a final volume \(V_2\) is given by the integral of pressure with respect to volume, which can be expressed as: \( W = \int_{V_1}^{V_2} P dV \) In this exercise, the temperature is held constant. This means that we have an isothermal process, and we can use the ideal gas law to solve for work done.
02

Apply the ideal gas law

The ideal gas law is given by: \( PV = nRT \) Here, P: pressure V: volume n: number of moles R: the ideal gas constant T: temperature Now, we have to express the pressure in terms of volume using the ideal gas equation. We can rewrite the equation as \(P = \dfrac{nRT}{V}\) since the number of moles, the ideal gas constant, and temperature are constants in this problem.
03

Determine the work done by the gas

Substitute the rewritten expression for pressure from Step 2 into the integral from Step 1: \(W = \int_{V_1}^{V_2} \dfrac{nRT}{V} dV\) To evaluate this integral, we will use the properties of natural logarithms. The integral of \( \dfrac{1}{V} dV \) is the natural logarithm, \( ln(V)\). Therefore, \(W = nRT \int_{V_1}^{V_2} \dfrac{1}{V} dV = nRT [ \ln(V_2) - \ln(V_1) ]\) Now, apply the logarithmic property \( ln(a) - ln(b) = ln(\dfrac{a}{b})\): \(W = nRT \ln{\dfrac{V_2}{V_1}}\)
04

Match the result with the correct answer choice

Our final expression for the work done by the gas is: \(W = nRT \ln{\dfrac{V_2}{V_1}}\) where n is the number of moles, R is the ideal gas constant, T is the temperature, and \(V_1\) and \(V_2\) are the initial and final volumes, respectively. This matches answer choice (B), which is the correct answer: (B) \(\mu \operatorname{RTln}\left\\{\mathrm{V}_{2} / \mathrm{V}_{1}\right\\}\)

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Most popular questions from this chapter

A diatomic gas initially at \(18^{\circ} \mathrm{C}\) is Compressed adiabatically to one eighth of its original volume. The temperature after Compression will be (A) \(10^{\circ} \mathrm{C}\) (B) \(668 \mathrm{~K}\) (C) \(887^{\circ} \mathrm{C}\) (D) \(144^{\circ} \mathrm{C}\)

A thermodynamic Process in which temperature \(\mathrm{T}\) of the system remains constant throughout Variable \(\mathrm{P}\) and \(\mathrm{V}\) may Change is called (A) Isothermal Process (B) Isochoric Process (C) Isobasic Process (D) None of this

The temperature on Celsius scale is \(25^{\circ} \mathrm{C}\). What is the corresponding temperature on the Fahrenheit Scale? (A) \(40^{\circ} \mathrm{F}\) (B) \(45^{\circ} \mathrm{F}\) (C) \(50^{\circ} \mathrm{F}\) (D) \(77^{\circ} \mathrm{F}\)

One \(\mathrm{kg}\) of adiatomic gas is at a pressure of $5 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ The density of the gas is \(\left\\{(5 \mathrm{~kg}) / \mathrm{m}^{3}\right\\}\) what is the energy of the gas due to its thermal motion ? (A) \(2.5 \times 10^{5} \mathrm{~J}\) (B) \(3.5 \times 10^{5} \mathrm{~J}\) (C) \(4.5 \times 10^{5} \mathrm{~J}\) (D) \(1.5 \times 10^{5} \mathrm{~J}\)

One mole of \(\mathrm{O}_{2}\) gas having a Volume equal to \(22.4\) liter at \(\mathrm{O}\) \(c\) and 1 atmospheric Pressure is Compressed isothermally so that its volume reduces to \(11.2\) liters. The work done in this Process is (A) \(1672.4 \mathrm{~J}\) (B) \(-1728 \mathrm{~J}\) (C) \(1728 \mathrm{~J}\) (D) \(-1572.4 \mathrm{~J}\)
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