Question

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

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

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: \(V_f = 3 \times V_i = 3 \times 2.00 = 6.00 \, m^3\).

Write the pressure function

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

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: $$W = \left[ \frac{(4.00 \, N/m^{11}) \, V^4}{4} \right]_{2.00}^{6.00}$$

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 \(1280 \, N \cdot m\).

Key concepts

Work Done by Gas

In thermodynamics, when a gas expands or compresses, it performs work. Work done by a gas is often calculated using the integral of pressure with respect to volume over the path of expansion or compression. Imagine pushing a gas-filled piston. The force exerted by the gas moving the piston does work, and this work can be represented by the area under the pressure-volume curve in a P-V diagram.

• The formula for work done by a gas is given by \[ W = \int_{V_i}^{V_f} P \, dV \]
• This integral sums the infinitesimal bits of work done as the volume changes from the initial volume \(V_i\) to the final volume \(V_f\).
• For our specific problem, the work done by the gas is calculated as \( 1280 \, N \cdot m \).

It's important to understand that the work done by the gas depends on how the pressure changes with volume, which is described by the pressure-volume relationship.

Pressure-Volume Relationship

The pressure-volume relationship for a gas describes how the pressure changes as the volume changes. In many thermodynamic processes, this relationship is a crucial part of calculating work.

• In our exercise, the pressure \( P \) of the gas is given by a power-law relationship: \( P = \alpha V^3 \).
• Here, \( \alpha \) is a constant with a dimension of \( 4.00 \, N/m^{11} \).
• This means that pressure increases rapidly as volume increases, following a cubic relation to the volume, which is not linear.

Understanding this relationship is vital. It tells us how hard the gas "pushes" back as it expands. Knowing \( P \) allows us to integrate and find the work done. This relationship is not typical for every gas and situation, showcasing the diverse behaviors gases can exhibit.

Integration in Physics

Integration is a mathematical method that finds the whole from knowing parts, and it's pivotal in physics, especially in calculating work.

• In the context of our problem, integration allows us to sum an infinite number of infinitesimal work elements as the gas changes volume.
• The integral \( \int_{V_i}^{V_f} \alpha V^3 \, dV \) evaluates the total work done by the expanding gas between initial \( V_i = 2.00 \, m^3 \) and final volumes \( V_f = 6.00 \, m^3 \).
• By solving the integral, we determine the net effect of pressure changes with volume, aiding in deeper understanding of the process.

Integrating in physics isn't just about calculating work; it's a tool for many physical concepts, including finding areas, volumes, and solving differential equations. Mastering integration helps solve complex problems by breaking them into manageable parts.