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Chapter 18: Problem 24

A gas has an initial volume of \(2.00 \mathrm{~m}^{3}\). It is expanded to three times its original volume through a process for which \(P=\alpha V^{3},\) with \(\alpha=4.00 \mathrm{~N} / \mathrm{m}^{11} .\) How much work is done by the expanding gas?

Short Answer

Expert verified
Question: Calculate the work done by an expanding gas whose pressure function is \(P = 4.00 \, N/m^{11} \, V^3\), when its volume expands from \(2.00 \, m^3\) to three times its initial volume. Answer: The work done by the expanding gas is \(1280 \, N \cdot m\).

Step by step solution

01

Determine the initial and final volumes

The initial volume, \(V_i\), is given as \(2.00 \, m^3\). The final volume, \(V_f\), is three times the initial volume: \(V_f = 3 \times V_i = 3 \times 2.00 = 6.00 \, m^3\).
02

Write the pressure function

The pressure function is given as \(P = \alpha V^3\), with \(\alpha = 4.00 \, N/m^{11}\).
03

Calculate the work done by the gas

The work done by the gas can be calculated by integrating the pressure function with respect to volume, from \(V_i\) to \(V_f\): $$W = \int_{V_i}^{V_f} P \, dV = \int_{2.00}^{6.00} \alpha V^3 \, dV$$ Next, substitute the value of \(\alpha\) into the equation: $$W = \int_{2.00}^{6.00} (4.00 \, N/m^{11}) \, V^3 \, dV$$ Now, integrate the equation with respect to volume: $$W = \left[ \frac{(4.00 \, N/m^{11}) \, V^4}{4} \right]_{2.00}^{6.00}$$
04

Evaluate the integral

Now, evaluate the integral using the limits of integration: $$W = \frac{(4.00 \, N/m^{11}) \, (6.00)^4}{4} - \frac{(4.00 \, N/m^{11}) \, (2.00)^4}{4}$$ Now, calculate \(W\): $$W = (4.00 \, N/m^{11}) \left[\frac{(6.00)^4-(2.00)^4}{4} \right]$$ $$W = (4.00 \, N/m^{11}) \left[\frac{(1296 -16)}{4} \right]$$ $$W = (4.00 \, N/m^{11}) \left[\frac{1280}{4} \right]$$ $$W = (4.00 \, N/m^{11}) \times 320 \, m^4$$ $$W = 1280 \, N \cdot m $$ The work done by the expanding gas is \(1280 \, N \cdot m\).

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