Question

Two moles of an ideal gas are expanded isothermally from $15\ \text{dm}^3$ to $20\ \text{dm}^3$. If the amount of work done is $-6\ \text{dm}^3\ \text{bar}$, find external pressure needed to obtain this work.

• A. $1.2 \times 10^5\ \text{Pa}$
• B. $3.2\ \text{Pa}$
• C. $8.1 \times 10^4\ \text{Pa}$
• D. $2.4\ \text{Pa}$

Hint:

Always keep track of signs in thermodynamics! Negative work explicitly confirms that work is being done by the system during an expansion ($V_2 > V_1$).
Since pressure must always be a positive physical value, your negative signs should cancel out perfectly during calculation: $P = \frac{|W|}{\Delta V}$.
mathP = 65 = 1.2
bar = 1.2 10^5
Pamath.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to calculate the constant external pressure required to perform an isothermal expansion of an ideal gas. We are given the initial volume, final volume, and total work done.

Step 2: Key Formula or Approach:
The expansion work done against a constant external pressure ($P_{\text{ext}}$) is given by the irreversible thermodynamic work formula:
$$W = -P_{\text{ext}} \Delta V$$
Where $\Delta V = V_2 - V_1$ is the change in system volume.
To match the standard SI unit options presented in Pascals (Pa), we can convert the computed pressure from bar using the conversion factor: $1\ \text{bar} = 10^5\ \text{Pa}$.

Step 3: Detailed Explanation:
Let's list the given values from the problem statement:
Number of moles ($n$) = $2$
Initial Volume ($V_1$) = $15\ \text{dm}^3$
Final Volume ($V_2$) = $20\ \text{dm}^3$
Work Done ($W$) = $-6\ \text{dm}^3\ \text{bar}$
First, calculate the volume change ($\Delta V$):
$$\Delta V = V_2 - V_1 = 20\ \text{dm}^3 - 15\ \text{dm}^3 = 5\ \text{dm}^3$$
Now, substitute $W$ and $\Delta V$ back into the core pressure equation:
$$-6\ \text{dm}^3\ \text{bar} = -P_{\text{ext}} \times 5\ \text{dm}^3$$
Isolate the constant external pressure variable ($P_{\text{ext}}$):
$$P_{\text{ext}} = \frac{-6}{-5}\ \text{bar} = 1.2\ \text{bar}$$
Finally, convert the bar pressure metric into standard Pascals:
$$P_{\text{ext}} = 1.2 \times 10^5\ \text{Pa}$$

Step 4: Final Answer:
The external pressure required is $1.2 \times 10^5\ \text{Pa}$, which corresponds precisely to option (A).