Question

What is the work done when a gas is compressed from $2.5 \times 10^{-2}\text{ m}^3$ to $1.3 \times 10^{-2}\text{ m}^3$ at constant external pressure of $4.05\text{ bar}$?

• A. $4050\text{ J}$
• B. $4400\text{ J}$
• C. $4200\text{ J}$
• D. $4860\text{ J}$

Hint:

Compression means work is done ON the system, so $W$ must be positive. Expansion means work is done BY the system, so $W$ must be negative. You can often eliminate half the options just by checking the sign!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a process where a gas is compressed from an initial volume to a final volume under a constant external pressure. We need to calculate the mechanical work done during this process in Joules.

Step 2: Key Formula or Approach:
The irreversible pressure-volume work ($W$) done against a constant external pressure ($P_{ext}$) is given by:
$$W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$$
To get the answer in Joules, pressure must be in Pascals ($\text{N/m}^2$) and volume must be in cubic meters ($\text{m}^3$).
Remember the conversion: $1\text{ bar} = 10^5\text{ Pa}$.

Step 3: Detailed Explanation:
First, identify the given variables:
Initial volume, $V_1 = 2.5 \times 10^{-2}\text{ m}^3$
Final volume, $V_2 = 1.3 \times 10^{-2}\text{ m}^3$
External pressure, $P_{ext} = 4.05\text{ bar} = 4.05 \times 10^5\text{ Pa}$
Next, calculate the change in volume ($\Delta V$):
$$\Delta V = V_2 - V_1 = (1.3 \times 10^{-2} - 2.5 \times 10^{-2})\text{ m}^3 = -1.2 \times 10^{-2}\text{ m}^3$$ .
Now, substitute the values into the work formula:
$$W = -(4.05 \times 10^5) \times (-1.2 \times 10^{-2})$$
$$W = + (4.05 \times 1.2) \times 10^3$$
$$W = 4.86 \times 10^3\text{ J}$$
$$W = 4860\text{ J}$$
The work is positive because it is a compression process (work is done on the system).

Step 4: Final Answer:
The work done is $4860\text{ J}$, which corresponds to option (D).