The problem states that a gas is heated at constant pressure. We are asked to determine the work done by the gas, given that the volume increases from $1 \, \text{m}^3$ to $1.5 \, \text{m}^3$ and the pressure is $100 \, \text{kPa}$.

Solution:

At constant pressure, the work $W$ done by the gas is calculated using the formula:

$W = P \Delta V$

Where:

• $P = 100 \, \text{kPa} = 100 \times 10^3 \, \text{Pa}$,
• $\Delta V = V_2 - V_1 = 1.5 \, \text{m}^3 - 1 \, \text{m}^3 = 0.5 \, \text{m}^3$.

Substituting the values:

$W = (100 \times 10^3) \times 0.5$

$W = 50,000 \, \text{J} = 50 \, \text{kJ}.$

Thus, the correct answer is:

b. 50 kJ

Would you like a more detailed explanation or additional insights?

Here are 5 related questions you might find interesting:

1. What is the physical significance of work done by a gas in a thermodynamic system?
2. How does the work done change if the pressure were not constant?
3. What are the units and conversion factors for pressure and volume in this context?
4. How would you calculate work done if the process were isothermal instead of constant pressure?
5. Can this formula $W = P \Delta V$ be applied to liquids? Why or why not?

Tip: Always remember to ensure consistent units (e.g., converting $\text{kPa}$ to $\text{Pa}$) before performing calculations.