NEET Physics Isothermal & Adiabatic Processes MCQ Question
Calculate the work done (in Joules) when 2 moles of an ideal gas expand isothermally and reversibly from a volume of 10 L to 20 L at a temperature of 300 K.
3457 J
3894 J
4157 J
4392 J
Correct Answer
Detailed Explanation
Using the formula for work done during isothermal expansion, w = -2.303 nRT log(Vf/Vi), we substitute n = 2, R = 8.314 J/mol·K, T = 300 K, Vi = 10 L, Vf = 20 L to find w = -2.303 * 2 * 8.314 * 300 * log(2) ≈ -3894 J.
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