\(2\) 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.
Step 1: Change the temperature to kelvin right away.
$T=27^{\circ}\text{C}=27+273=300$ K. Working in kelvin from the start avoids mistakes later.
Step 2: Recall why the work formula has a log term.
For an ideal gas, pressure is $p=\dfrac{nRT}{V}$. At constant $T$, the reversible work is the area under the $p$-$V$ curve,
\[ W=\int_{V_1}^{V_2}p\,dV=nRT\int_{V_1}^{V_2}\frac{dV}{V}=nRT\ln\left(\frac{V_2}{V_1}\right) \]
Step 3: Combine $n$, $R$ and $T$ into a single number first.
\[ nRT=2\times8.314\times300=4988.4\ \text{J} \]
Step 4: Handle the volume ratio and its log separately.
\[ \frac{V_2}{V_1}=\frac{20\ \text{L}}{10\ \text{L}}=2,\qquad \ln2=0.6931 \]
Step 5: Multiply the two pieces.
\[ W=4988.4\times0.6931=3457.70\ \text{J} \]
This matches the direct substitution method and confirms the work magnitude is about $3457.70$ J, which falls inside the given band of $3440.00$ to $3492.00$ J.
\[ \boxed{W\approx3457.70\ \text{J}} \]
