Two moles of an ideal gas are expanded irreversibly and isothermally at $37^{\circ} C$ until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.
$Δ S_{t o t a l} (system + surrounding)$ is $(R = \frac{25}{3} J/molK, ln 2=0.7, ln 3 = 1.1)$ :

- 0.667 J/K

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0.667 J/K

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22.7 J/K

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Solution

The correct option is B 0.667 J/K
$Δ S = \frac{q}{T}$
For surrounding:
$q = - 3.41 k J$
$Δ S = - \frac{3.41 \times 1000}{37 + 273} = - 11 J K^{- 1}$
For a process which is irreversible and isothermal:
$Δ S = C_{v} l n \frac{T_{f}}{T_{i}} + n R l n \frac{V_{f}}{V_{i}}$
$Δ S = n R l n \frac{V_{f}}{V_{i}}$
$Δ S = 2 \times 8.314 l n \frac{2 V}{V} = 11.667$
$Δ S_{t o t a l} = Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g}$
$= 0.667$

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Two moles of an ideal gas are expanded irreversibly and isothermally at 37∘ C until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings. ΔStotal(system + surrounding) is (R=253 J/molK, ln 2=0.7, ln 3 = 1.1):

The correct option is B 0.667 J/KΔS=qT For surrounding: q=−3.41kJ ΔS=−3.41×100037+273=−11 JK−1 For a process which is irreversible and isothermal: ΔS=CvlnTfTi+nRlnVfVi ΔS=nRlnVfVi ΔS=2×8.314ln2VV=11.667 ΔStotal=ΔSsystem+ΔSsurrounding =0.667

Two moles of an ideal gas are expanded irreversibly and isothermally at $37^{\circ} C$ until the volume is doubled, and 3.41 kJ heat is absorbed from the surroundings.
$Δ S_{t o t a l} (system + surrounding)$ is $(R = \frac{25}{3} J/molK, ln 2=0.7, ln 3 = 1.1)$ :