The entropy (in JK -1 ) of mixing of 100 g N 2 O at 120° C and 1.5 bar with 100 g of CO 2 at 120° C and 1.5 bar is ( log 2=0.300)

Question

The entropy (in JK -1 ) of mixing of 100 g N 2 O at 120° C and 1.5 bar with 100 g of CO 2 at 120° C and 1.5 bar is ( log 2=0.300) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ S text mixing =-[ n 1 R ln X 1+ n 2 R ln X 2] =-[(100/44) × 8.315 ln (1/2)+(100/44) × 8.315 ln (1/2)] =+(100/44) × 8.135 × 2.303 log 2=26.19 ( J / K )

Q. The entropy (in $J K^{- 1}$ ) of mixing of $100 g N_{2} O$ at $12 0^{\circ} C$ and $1.5$ bar with $100 g$ of $C O_{2}$ at $12 0^{\circ} C$ and $1.5$ bar is $(lo g 2 = 0.300)$

$Δ S_{mixing} = - [n_{1} R ln X_{1} + n_{2} R ln X_{2}]$
$= - [\frac{100}{44} \times 8.315 ln \frac{1}{2} + \frac{100}{44} \times 8.315 ln \frac{1}{2}]$
$= + \frac{100}{44} \times 8.135 \times 2.303 lo g 2 = 26.19 \frac{J}{K}$

Q. The entropy (in JK−1 ) of mixing of 100gN2​O at 120∘C and 1.5 bar with 100g of CO2​ at 120∘C and 1.5 bar is (log2=0.300)

ΔSmixing ​=−[n1​RlnX1​+n2​RlnX2​] =−[44100​×8.315ln21​+44100​×8.315ln21​] =+44100​×8.135×2.303log2=26.19KJ​

Q. The entropy (in $J K^{- 1}$ ) of mixing of $100 g N_{2} O$ at $12 0^{\circ} C$ and $1.5$ bar with $100 g$ of $C O_{2}$ at $12 0^{\circ} C$ and $1.5$ bar is $(lo g 2 = 0.300)$

$Δ S_{mixing} = - [n_{1} R ln X_{1} + n_{2} R ln X_{2}]$
$= - [\frac{100}{44} \times 8.315 ln \frac{1}{2} + \frac{100}{44} \times 8.315 ln \frac{1}{2}]$
$= + \frac{100}{44} \times 8.135 \times 2.303 lo g 2 = 26.19 \frac{J}{K}$