Two mole of an ideal gas is expanded irreversibly and isothermally at 37 -C until its volume is doubled and 3-41 kJ heat is absorbed from surrounding- - Stotal system - surrounding is -

The correct option is A 0.52 J/K Δ S_system = nbsp;nRInV_2/V_1 = 2 × R × In2 = 11.52 J/k Δ S_surrounding = 3.41 × 1000/310 = 11 J/K Δ S_ ...

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Standard XII

Chemistry

Entropy Change in Irreversible Process

Two mole of a...

Question

Two mole of an ideal gas is expanded irreversibly and isothermally at 37 $^{\circ}$ C until its volume is doubled and 3.41 kJ heat is absorbed from surrounding. $Δ S_{t o t a l}$ (system + surrounding) is :

0

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0.52 J / K

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22.52 J / K

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none of the above

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Solution

The correct option is A $0.52 J / K$
$Δ S_{s y s t e m} = n R I n \frac{V_{2}}{V_{1}} = 2 \times R \times I n 2 = 11.52 J / k$
$Δ S_{s u r r o u n d i n g} = \frac{3.41 \times 1000}{310} = - 11 J / K$
$Δ S_{t o t a l} = + 11.52 - 11 = + 0.52 J / K$

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Two mole of an ideal gas is expanded irreversibly and isothermally at 37∘C until its volume is doubled and 3.41 kJ heat is absorbed from surrounding. ΔStotal (system + surrounding) is :

The correct option is A 0.52J/KΔSsystem=nRInV2V1=2×R×In2=11.52J/kΔSsurrounding=3.41×1000310=−11J/KΔStotal=+11.52−11=+0.52J/K

Two mole of an ideal gas is expanded irreversibly and isothermally at 37 $^{\circ}$ C until its volume is doubled and 3.41 kJ heat is absorbed from surrounding. $Δ S_{t o t a l}$ (system + surrounding) is :

The correct option is A $0.52 J / K$
$Δ S_{s y s t e m} = n R I n \frac{V_{2}}{V_{1}} = 2 \times R \times I n 2 = 11.52 J / k$
$Δ S_{s u r r o u n d i n g} = \frac{3.41 \times 1000}{310} = - 11 J / K$
$Δ S_{t o t a l} = + 11.52 - 11 = + 0.52 J / K$