Question

The work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 1L to 10L at 300 K is (R = 0.0083 kJK mol^-1)

• A. 0.115 kJ
• B. 11.5 kJ
• C. 58.5 kJ
• D. 5.8 kJ

Answer 1

The Correct Option is B

Correct Answer: 11.5 kJ 

Explanation:
For isothermal, reversible expansion of an ideal gas, the work done is given by:
W = -nRT ln(Vf/Vi)

Where:
• n = 2 mol
• R = 0.0083 kJ·K⁻¹·mol⁻¹
• T = 300 K
• Vi = 1 L
• Vf = 10 L

Substituting the values:
\(W = -2 \times 0.0083 \times 300 \times \ln\left(\frac{10}{1}\right)\)
\(W = -4.98 \times \ln(10)\)
Since \(\ln(10) \approx 2.303\),
\(W = -4.98 \times 2.303 \approx -11.47 \text{ kJ}\)

Work done by the system is positive in expansion:
So, work done = 11.5 kJ.

Answer 2

The work done during an isothermal expansion of an ideal gas can be calculated using the equation:
\(W = -nRT \ln \left( \frac{V_f}{V_i} \right)\)
Given:
n = 2 moles
R = 0.0083 kJ/mol K
T = 300 K 
\(V_i = 1 \, \text{L}\)
\(V_f = 10 \, \text{L}\)
Plugging in the values, we have:
\(W = -2 \times (0.0083 \, \text{kJ/mol K}) \times 300 \, \text{K} \times \ln \left( \frac{10}{1} \right)\)
\(W \approx -2 \times 0.0083 \, \text{kJ} \times 300 \times \ln(10)\)
\(W ≈ -49.8 ln(10)\)
Using a calculator to evaluate the natural logarithm of 10 and multiplying by -49.8, we find:
\(W \approx -49.8 \times 2.3026\)
\(W ≈ -114.80748 ≈ -115 kJ\) (rounded to three significant figures)
Since work is a transfer of energy, it is conventionally expressed as a positive value. 
Therefore, the work done during the isothermal expansion is approximately (B) 11.5 kJ.