Verified Solution • 2025–26 Session moderate

QNEET-2016-Ch-1108: NEET 2016 Chemistry - Class 11 NEET-2016-Ch-1108 Solution: For a sample of perfect gas when its pressure is changed…

For a sample of perfect gas when its pressure is changed isothermally from pip_i to pfp_f the entropy change is given by ΔS=nRln⁡(pfpi)\Delta S = nR\ln(\frac{p_f}{p_i}) ΔS=nRln⁡(pipf)\Delta S = nR\ln(\frac{p_i}{p_f}) ΔS=nRTln⁡(pfpi)\Delta S = nRT\ln(\frac{p_f}{p_i}) ΔS=RTln⁡(pipf)\Delta S = RT\ln(\frac{p_i}{p_f})

Multiple Choice Options

  • A ΔS=nR ln⁡(pfpi)\Delta S = nR\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right) _ S = n R ln ( p i _ p f _ _ )
  • B ΔS=nR ln⁡(pipf)\Delta S = nR\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right) _ S = n R ln ( p f _ p i _ _ )
  • C ΔS=nRT ln⁡(pfpi)\Delta S = nRT\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right) _ S = n RT ln ( p i _ p f _ _ )
  • D ΔS=RT ln⁡(pipf)\Delta S = RT\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right) _ S = RT ln ( p f _ p i _ _ )

Step-by-Step Verified Solution

For an ideal gas undergoing reversible expansion, when temperature changes from T i to T f and pressure changes from P i to P f . _ S = nC p ln T f T i + nR ln P i P f For an isothermal process, T i = T f so, ln 1 = 0 _ _ S = nR ln P i P f

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