Question

A gas is taken through the cycle $ A \to B \to C \to A $, as shown in the figure. What is the net work done by the gas?

• A. 2000 J
• B. 1000 J
• C. Zero
• D. -2000 J

Hint:

To calculate the net work done by a gas in a cyclic process, find the area enclosed by the cycle on the PV diagram. The area represents the work done by the gas.

Answer 1

The Correct Option is B

To solve the problem, we need to calculate the net work done by a gas using the area enclosed in the pressure-volume (p-V) curve.

1. Understanding the Problem:
The net work done by a gas is given by the area enclosed in the $p$-$V$ curve. In this case, we need to calculate the area of triangle $\triangle ABC$.

2. Formula for the Area of a Triangle:
The formula for the area of a triangle is:

$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $

3. Given Values:
- Base of the triangle: $5 \times 10^{-3} \, \text{m}^3$
- Height of the triangle: $4 \times 10^5 \, \text{Pa}$

4. Calculate the Net Work Done:
Substitute the given values into the area formula:

$ W_{\text{net}} = \frac{1}{2} \times \text{base} \times \text{height} $
$ W_{\text{net}} = \frac{1}{2} \times (5 \times 10^{-3}) \times (4 \times 10^5) $

5. Simplify the Expression:
First, multiply the constants:

$ W_{\text{net}} = \frac{1}{2} \times 5 \times 4 \times 10^{-3} \times 10^5 $
$ W_{\text{net}} = \frac{1}{2} \times 20 \times 10^{2} $
$ W_{\text{net}} = 10 \times 10^2 $
$ W_{\text{net}} = 10^3 \, \text{J} $

6. Final Answer:
The net work done by the gas is $1000 \, \text{J}$.

Answer 2

Step 1: Understanding the Problem

The net work done by a gas is given by the area enclosed in the $ p $-$ V $ (pressure-volume) curve. In this case, the area of interest is the area of triangle $ \triangle ABC $.

Step 2: Formula for the Area of a Triangle

The formula for the area of a triangle is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Step 3: Given Values

• Base of the triangle: $ 5 \times 10^{-3} \, \text{m}^3 $
• Height of the triangle: $ 4 \times 10^5 \, \text{Pa} $

Step 4: Calculate the Net Work Done

Substitute the given values into the formula for the area: \[ W_{\text{net}} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ W_{\text{net}} = \frac{1}{2} \times (5 \times 10^{-3}) \times (4 \times 10^5) \]

Step 5: Simplify the Expression

First, multiply the constants: \[ W_{\text{net}} = \frac{1}{2} \times 5 \times 4 \times 10^{-3} \times 10^5 \] \[ W_{\text{net}} = \frac{1}{2} \times 20 \times 10^{2} \] \[ W_{\text{net}} = 10 \times 10^2 \] \[ W_{\text{net}} = 10^3 \, \text{J} \]

Step 6: Final Answer

The net work done by the gas is: \[ \boxed{1000 \, \text{J}} \]