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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

In an isothermal reversible expansion, if the volume of \(96 \mathrm{~J}\) of oxygen at \(27^{\circ} \mathrm{C}\) is increased from 70 liter to 140 liter, then the work done by the gas will be (A) \(300 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (B) \(81 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (C) \(2.3 \times 900 \mathrm{R} \log _{10} 2\) (D) \(100 \mathrm{R} \log _{10}^{(2)}\)

In a container of negligible heat capacity, \(200 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) and \(100 \mathrm{~g}\) steam at \(100^{\circ} \mathrm{C}\) are added to \(200 \mathrm{~g}\) of water that has temperature $55^{\circ} \mathrm{C}$. Assume no heat is lost to the surroundings and the pressure in the container is constant atm. What is the final temperature the System? (A) \(72^{\circ} \mathrm{C}\) (B) \(48^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(94^{\circ} \mathrm{C}\)

Water of volume 2 liter in a container is heated with a coil of $1 \mathrm{kw}\( at \)27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at the rate of \(160(\mathrm{~J} / \mathrm{S}) .\) In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to $77^{\circ} \mathrm{C}\(. Specific heat of water is \)4.2\\{(\mathrm{KJ}) /(\mathrm{Kg})\\}$ (A) \(7 \mathrm{~min}\) (B) \(6 \min 2 \mathrm{~s}\) (C) \(14 \mathrm{~min}\) (D) \(8 \min 20 \mathrm{~S}\)

For which combination of working temperatures the efficiency of Carnot's engine is highest. (A) \(80 \mathrm{~K}, 60 \mathrm{~K}\) (B) \(100 \mathrm{~K}, 80 \mathrm{~K}\) (C) \(60 \mathrm{~K}, 40 \mathrm{~K}\) (D) \(40 \mathrm{~K}, 20 \mathrm{~K}\)

Which of the following is not a thermodynamical function. (A) Enthalpy (B) Work done (C) Gibb's energy (D) Internal energy
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