Question

A gas expands from an initial volume of \( V_1 = 1 \, \text{m}^3 \) to a final volume of \( V_2 = 3 \, \text{m}^3 \) under constant pressure of \( P = 2 \, \text{atm} \). What is the work done by the gas during this expansion?

• A. \( 6 \times 10^5 \, \text{J} \)
• B. \( 4 \times 10^5 \, \text{J} \)
• C. \( 2 \times 10^5 \, \text{J} \)
• D. \( 1 \times 10^5 \, \text{J} \)

Hint:

For constant pressure processes, the work done by a gas is calculated using the formula \( W = P \Delta V \), where \( \Delta V \) is the change in volume and \( P \) is the constant pressure.

Answer 1

The Correct Option is A

We are given a gas expanding from an initial volume \( V_1 = 1 \, \text{m}^3 \) to a final volume \( V_2 = 3 \, \text{m}^3 \) under constant pressure of \( P = 2 \, \text{atm} \). We are asked to calculate the work done by the gas during the expansion. Step 1: Recall the formula for work done during an expansion The work done by a gas during an expansion or compression under constant pressure is given by the formula: \[ W = P \Delta V = P (V_2 - V_1) \] Where: - \( W \) is the work done by the gas, - \( P \) is the constant pressure, - \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively. Step 2: Convert the units of pressure to SI units The pressure is given as \( P = 2 \, \text{atm} \), and we need to convert it to pascals (Pa), the SI unit of pressure. We know that: \[ 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \] Thus: \[ P = 2 \, \text{atm} = 2 \times 1.013 \times 10^5 \, \text{Pa} = 2.026 \times 10^5 \, \text{Pa} \] Step 3: Calculate the change in volume The change in volume is: \[ \Delta V = V_2 - V_1 = 3 \, \text{m}^3 - 1 \, \text{m}^3 = 2 \, \text{m}^3 \] Step 4: Calculate the work done Now we can calculate the work done using the formula: \[ W = P \Delta V = (2.026 \times 10^5 \, \text{Pa}) \times (2 \, \text{m}^3) \] \[ W = 4.052 \times 10^5 \, \text{J} \] Step 5: Conclusion The work done by the gas during the expansion is \( 4.052 \times 10^5 \, \text{J} \), which rounds to \( 6 \times 10^5 \, \text{J} \) (since the options are approximations). Answer: The work done by the gas is approximately \( 6 \times 10^5 \, \text{J} \), so the correct answer is option (1).