Question

\(2\) moles of an ideal gas are reversibly expanded from \(10\) litre to \(20\) litre under isothermal conditions. The universal gas constant \(R = 8.314\ \text{J/mol-K}\). If the temperature of the gas is \(27^{\circ}\text{C}\), find the magnitude of the work done in the process (rounded off to two decimal places), in Joules.

Hint:

Use \(W=nRT\ln(V_2/V_1)\) with \(T\) in kelvin.

Answer 1

Correct Answer: 3440

Step 1: Recall the work done in a reversible isothermal expansion.
For \(n\) moles of an ideal gas expanding reversibly at constant temperature \(T\) from volume \(V_1\) to \(V_2\), the work done by the gas is
\[ W = nRT\ln\left(\frac{V_2}{V_1}\right) \]
This comes from integrating the ideal gas pressure \(p = nRT/V\) over the volume change, since \(T\) stays fixed.

Step 2: Write down the given values.
\[ n = 2\ \text{mol}, \quad R = 8.314\ \text{J/mol-K}, \quad V_1 = 10\ \text{L}, \quad V_2 = 20\ \text{L} \]
The temperature is \(27^{\circ}\text{C}\), which in kelvin is
\[ T = 27 + 273 = 300\ \text{K} \]

Step 3: Find the volume ratio.
\[ \frac{V_2}{V_1} = \frac{20}{10} = 2 \]
Since the work depends only on the ratio of volumes, there is no need to convert litres to \(\text{m}^3\).

Step 4: Compute the natural log of the ratio.
\[ \ln(2) = 0.6931 \]

Step 5: Substitute all values into the work formula.
First find \(nRT\):
\[ nRT = 2 \times 8.314 \times 300 = 4988.4\ \text{J} \]
Then multiply by \(\ln(2)\):
\[ W = 4988.4 \times 0.6931 = 3457.70\ \text{J} \]

Final Answer:
The magnitude of the work done in the isothermal reversible expansion is about 3457.70 J.
\[ \boxed{W \approx 3457.70\ \text{J}} \]