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Chapter 12: Problem 25

\(n\) moles of a gas expands from volume \(V_{1}\) to \(V_{2}\) at constant temperature \(T\). The work done by the gas is (A) \(n R T\left(\frac{v_{2}}{v_{1}}\right)\) (B) \(n R T\left(\frac{v_{2}}{v_{1}}-1\right)\) (C) \(n R T \ln \left(\frac{v_{1}}{v_{2}}\right)\) (D) None of these

Short Answer

Expert verified
The short answer is: (D) None of these. The correct expression for the work done by the gas in an isothermal expansion is \(W = -nRT\ln\left(\frac{V_{2}}{V_{1}}\right)\).

Step by step solution

01

Recall the Ideal Gas Law

Recall the Ideal Gas Law which states that for a given sample of gas, the product of its pressure (P), volume (V), and temperature (T) is proportional to the number of moles (n) of the gas: \[PV = nRT\] Here, R is the universal gas constant.
02

Recognize Isothermal Expansion

Isothermal Expansion occurs when the gas expands without changes in its temperature. In this exercise, the gas is expanding at a constant temperature T, which means it's an isothermal expansion. In isothermal expansion, the relationship between pressure and volume can be expressed as follows: \[P_{1}V_{1} = P_{2}V_{2}\]
03

Calculate the work done by the gas

To calculate the work (W) done by the gas during isothermal expansion, we can use the formula: \[W = -\int_{V_1}^{V_2} PdV\] By using the ideal gas law, we can express the pressure in terms of volume: \[P = \frac{nRT}{V}\] Now, substitute the pressure expression and integrate with respect to volume: \[W = -\int_{V_1}^{V_2} \frac{nRT}{V} dV\]
04

Integrate and simplify the expression

Integrate the expression with respect to V: \[W = -nRT \int_{V_1}^{V_2} \frac{1}{V} dV\] \[W = -nRT (\ln (V_2) - \ln (V_1))\] Using logarithmic properties: \[W = -nRT \ln\left(\frac{V_{2}}{V_{1}}\right)\] Now let us check the given options: (A) \(nRT\left(\frac{V_{2}}{V_{1}}\right)\) (B) \(nRT\left(\frac{V_{2}}{V_{1}}-1\right)\) (C) \(nRT \ln \left(\frac{V_{1}}{V_{2}}\right)\) (D) None of these
05

Identify the correct expression

Compare the derived expression with the given options. Our derived expression is \(W = -nRT \ln\left(\frac{V_{2}}{V_{1}}\right)\). We notice that this expression is the negative of option (C). Therefore, none of the options match the correct expression for the work done by the gas. The correct answer is: (D) None of these.

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Most popular questions from this chapter

A Carnot engine, whose efficiency is \(40 \%\), takes heat from a source maintained at a temperature of \(500 \mathrm{~K}\). It is desired to have an engine of efficiency \(60 \%\). Then, the intake temperature for the same exhaust (sink) temperature must be: [2012] (A) Efficiency of Carnot engine cannot be made larger than \(50 \%\) (B) \(1200 \mathrm{~K}\) (C) \(750 \mathrm{~K}\) (D) \(600 \mathrm{~K}\)

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