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Chapter 12: Problem 31

A vertical cylinder with a massless piston is filled with one mole of an ideal gas. The piston can move freely without friction. The piston is slowly raised so that the gas expands isothermally at temperature \(300 \mathrm{~K}\). The amount of work done in increasing the volume two times is \(R=\frac{25}{3} \mathrm{~J} / \mathrm{mol} / \mathrm{K}, \log _{\mathrm{e}} 2=0.7\) (A) \(1750 \mathrm{~J}\) (B) \(2500 \mathrm{~J}\) (C) \(750 \mathrm{~J}\) (D) \(4250 \mathrm{~J}\)

Short Answer

Expert verified
The amount of work done in increasing the volume two times is (B) \(2500\mathrm{~J}\).

Step by step solution

01

Identify given values

From the exercise, we have the following values: - n = 1 mole (since it is given that there is one mole of an ideal gas) - T = 300 K (the temperature remains constant throughout the process) - R = 25/3 J/mol/K (the gas constant) - ln(2) = 0.7 (natural logarithm of 2)
02

Set up the formula for work done in an isothermal process

We will use the following formula for work done in an isothermal process: \(W = nRT\ln(V_f / V_i)\) Here, W represents the work done, n is the number of moles, R is the gas constant, T is the temperature, and \(V_f\) and \(V_i\) are the final volume and initial volume respectively.
03

Find the final volume and initial volume

Since the volume increases two times, we can write the relationship between final and initial volumes as follows: \(V_f = 2V_i\) Now substitute this into the formula for work done: \(W = nRT\ln(2V_i / V_i)\)
04

Simplify the formula for work done

Now simplify the formula by cancelling out the initial volume terms in both the numerator and denominator: \(W = nRT\ln(2)\) Now we can plug in the values from the exercise into the formula and solve for W.
05

Calculate the work done

\(W = (1\,\text{mol})\cdot \frac{25}{3}\,\frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot (300\,\text{K})\cdot (0.7)\) \(W = \frac{25}{3}\text{ J/mol/K}\cdot 300\,\text{K} \cdot 0.7\) \(W = 25\cdot 100\,\text{J} \cdot 0.7\) \(W = 2500\,\text{J}\) So, the amount of work done in increasing the volume by two times is \(2500\,\text{J}\). Therefore, the correct answer is (B) \(2500\mathrm{~J}\).

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