Question

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50 points
please help Yrr

Answer 1

Where x₁ is the mole fraction of benzene ,
x₂ is the mole fraction of toluene , 
n is the total number of moles 
R is the universal gas constant e.g., R = 25/3 J/Kmol
T is the temperature .

Now, x₁ = (n₁)/(n₁ + n₂) = 6/(6 + 2.5) = 6/8.5 = 12/17 
x₂ = n₂/(n₁ + n₂) = 2.5/(6 + 2.5) = 5/17 

n = n₁ + n₂ = 6 + 2.5 = 8.5 
R = 25/3 J/Kmol
T = 298K 
∴= 8.5 × 25/3 × 298 [12/17ln(12/17) + 5/17ln(5/17)]
=12.5 × 298/3 [12ln(12/17) + 5ln(5/17) ] J 
= 12.5 × 298/3 × (-10.2985) = -12.787 kJ

Answer 2

Formula of Gibbs free energy of mixture is given by 

$\bold{\Delta_\text{mix}G= nRT(x_1\ln x_1 +x_2\ln x_2)}Δmix​G=nRT(x1​lnx1​+x2​lnx2​)$

Where x₁ is the mole fraction of benzene ,
x₂ is the mole fraction of toluene , 

n is the total number of moles 

R is the universal gas constant e.g., R = 25/3 J/Kmol

T is the temperature .

Now, x₁ = (n₁)/(n₁ + n₂) = 6/(6 + 2.5) = 6/8.5 = 12/17 
x₂ = n₂/(n₁ + n₂) = 2.5/(6 + 2.5) = 5/17 

Now, $\bold{\Delta G_{mix}= nRT(x_1 lnx_1+x_2 lnx_2)}ΔGmix​=nRT(x1​lnx1​+x2​lnx2​)$

n = n₁ + n₂ = 6 + 2.5 = 8.5 

R = 25/3 J/Kmol

T = 298K 

∴ $\bold{\Delta G_{mix}}ΔGmix​ = 8.5 × 25/3 × 298 [12/17ln(12/17) + 5/17ln(5/17)]$

=12.5 × 298/3 [12ln(12/17) + 5ln(5/17) ] J 

= 12.5 × 298/3 × (-10.2985)

= -12.787 kJ