(a) Calculate the magnitude, or absolute value, of the work done on the gas in this process. (Be careful with units. Your answer should be in joules.)

Steps to Solve

1. Convert pressures to consistent units

Convert the pressures from atmospheres (atm) to kilopascals (kPa):

• Initial pressure: $P_i = 1 , \text{atm} \times 101.3 , \text{kPa/atm} = 101.3 , \text{kPa}$
• Final pressure: $P_f = 2 , \text{atm} \times 101.3 , \text{kPa/atm} = 202.6 , \text{kPa}$

2. Identify the pressure-volume relationship

The work done by the gas during an isobaric (constant pressure) process is given by the formula: $$ W = P \Delta V $$ Where $\Delta V = V_f - V_i$, with:

• Initial volume, $V_i = 0.02 , \text{m}^3$
• Final volume, $V_f = 0.10 , \text{m}^3$

Calculate the change in volume: $$ \Delta V = 0.10 , \text{m}^3 - 0.02 , \text{m}^3 = 0.08 , \text{m}^3 $$

3. Calculate the work done on the gas

Use the average pressure for the work calculation: $$ P_{avg} = \frac{P_i + P_f}{2} = \frac{101.3 , \text{kPa} + 202.6 , \text{kPa}}{2} = 151.95 , \text{kPa} $$

Now calculate the work done: $$ W = P_{avg} \Delta V = 151.95 , \text{kPa} \times 0.08 , \text{m}^3 $$

4. Convert work into joules

Since $1 , \text{kPa} \cdot \text{m}^3 = 1000 , \text{J}$: $$ W = 151.95 \times 0.08 \times 1000 , \text{J} $$ Compute the final value to find the work done.