What is the free energy change - G when 1-0 mole of water at 100-C and 1 atm pressure-

1 mole of water at nbsp;100^∘C and 1 atm nbsp;→ Steam at 100^∘C and 1 atm n = 1 T_1=373 K P_1=1 atm T_2=373 K P_2=1 atm Δ G=nRT ln ( P_2P_1 ) =nRT ln( ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

What is the f...

Question

What is the free energy change $(Δ G)$ when 1.0 mole of water at $100^{\circ} C$ and 1 atm pressure?

80 cal

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620 cal

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Zero

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Solution

The correct option is D Zero
1 mole of water at $100^{\circ} C$ and 1 atm $\to$ Steam at
$100^{\circ} C$ and 1 atm
n = 1
$T_{1} = 373 K$
$P_{1} = 1 a t m$
$T_{2} = 373 K$
$P_{2} = 1 a t m$
$Δ G = n R T l n (\frac{P_{2}}{P_{1}})$
$= n R T l n (1)$
= 0 (ln 1 = 0)

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What is the free energy change (ΔG) when 1.0 mole of water at 100∘C and 1 atm pressure?

The correct option is D Zero1 mole of water at 100∘C and 1 atm → Steam at 100∘C and 1 atm n = 1 T1=373 K P1=1 atm T2=373 K P2=1 atm ΔG=nRTln (P2P1) =nRT ln(1) = 0 (ln 1 = 0)

What is the free energy change $(Δ G)$ when 1.0 mole of water at $100^{\circ} C$ and 1 atm pressure?

The correct option is D Zero
1 mole of water at $100^{\circ} C$ and 1 atm $\to$ Steam at
$100^{\circ} C$ and 1 atm
n = 1
$T_{1} = 373 K$
$P_{1} = 1 a t m$
$T_{2} = 373 K$
$P_{2} = 1 a t m$
$Δ G = n R T l n (\frac{P_{2}}{P_{1}})$
$= n R T l n (1)$
= 0 (ln 1 = 0)