Question

1 mole of an ideal gas expands isothermally and reversibly by decreasing pressure from $210\ \mathrm{kPa}$ to $105\ \mathrm{kPa}$ at $300\ \mathrm{K}$. What is the work done? ($R = 8.314\ \mathrm{J\ K}^{-1}\ \mathrm{mol}^{-1}$)

• A. $1960\ \mathrm{J}$
• B. $864\ \mathrm{J}$
• C. $1296\ \mathrm{J}$
• D. $1729\ \mathrm{J}$

Hint:

Whenever the volume doubles or the pressure is cut exactly in half ($P_1/P_2 = 2$) during an isothermal process, the core work term simplifies to $nRT \ln(2)$. Since $\ln(2) \approx 0.7$, a quick approximation gives $1 \times 8.3 \times 300 \times 0.7 \approx 1743\ \mathrm{J}$, pointing you straight to option (D).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem requires computing the mechanical work performed by $1\text{ mole}$ of an ideal gas undergoing an isothermal, reversible expansion process at a constant temperature of $300\ \mathrm{K}$ while its pressure decreases from an initial value of $210\ \mathrm{kPa}$ down to a final value of $105\ \mathrm{kPa}$.

Step 2: Key Formula or Approach:
The maximum work done ($W_{\text{max}}$) during a reversible, isothermal gas expansion can be determined using the integrated thermodynamic work equation in terms of pressures: $$W = -2.303 \cdot n \cdot R \cdot T \cdot \log_{10}\left(\frac{P_1}{P_2}\right)$$ Where: $n = 1\ \mathrm{mol}$ is the amount of substance. $R = 8.314\ \mathrm{J\ K}^{-1}\ \mathrm{mol}^{-1}$ is the universal gas constant. $T = 300\ \mathrm{K}$ is the absolute temperature. $P_1 = 210\ \mathrm{kPa}$ and $P_2 = 105\ \mathrm{kPa}$ are the initial and final pressures.
Note: If we consider work done by the gas as a positive magnitude of expansion work, the absolute work value is evaluated. Let's compute its magnitude to map it to the positive choices provided.

Step 3: Detailed Explanation:
Substitute the given values into the formula: $$W = -2.303 \times 1 \times 8.314 \times 300 \times \log_{10}\left(\frac{210}{105}\right)$$ Simplify the fractional pressure ratio inside the logarithm: $$\frac{210}{105} = 2$$ $$\log_{10}(2) \approx 0.3010$$ Now, multiply the remaining terms step by step: $$W = -2.303 \times 8.314 \times 300 \times 0.3010$$ Since $2.303 \times \log_{10}(2) = \ln(2) \approx 0.693$, we can simplify: $$W = -8.314 \times 300 \times 0.693$$ $$W = -2494.2 \times 0.693 \approx -1728.5\ \mathrm{J}$$ Taking the magnitude of the work done during this thermodynamic expansion process gives approximately $1729\ \mathrm{J}$.

Step 4: Final Answer:
The total work done is $1729\ \mathrm{J}$, which matches option (D).