NEET 2024 Chemistry Question 91 Solution
Question
The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is: (Given R = 2.0 cal K⁻¹ mol⁻¹)
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The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is: (Given R = 2.0 cal K⁻¹ mol⁻¹)
Gas EXPANDS from 20 atm to 10 atm -> Work is done BY the gas -> w is strictly NEGATIVE (-413.14 cal)! Eliminate positive options immediately!
In chemistry, work done on the system is positive, and work done by the system (such as during expansion against external pressure) is negative. For a reversible isothermal expansion of an ideal gas:
Expressing natural logarithm in terms of base-10 logarithm using the factor 2.303:
Given: - n = 1 mole - R = 2.0 cal K⁻¹ mol⁻¹ - T = 25 + 273.15 = 298.15 K (or 298 K) - P1 = 20 atm, P2 = 10 atm ⟹ P1/P2 = 2.0 - log10(2) = 0.30103
Multiplying the terms step by step: 2.303 × 2.0 = 4.606 4.606 × 298.15 = 1373.28 1373.28 × 0.30103 = 413.14 calories. With the negative sign for expansion work:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Physics sign convention; in chemical thermodynamics, expansion work is strictly negative.
Arbitrary incorrect value.
Internal energy change ΔU = 0 for isothermal process, but work w is non-zero.
Exact magnitude and correct IUPAC negative sign for expansion work.
Option (1) offers +413.14 cal. Remember that in chemistry, WORK DONE BY THE GAS IN EXPANSION IS ALWAYS NEGATIVE (w = -413.14 cal).
"Work done in a reversible isothermal expansion of an ideal gas is negative under IUPAC convention: w = -2.303 nRT log(P1/P2)."
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