Q1: A mixture of 10% C₆H₆ vapour in air at 25°C and 750 mm Hg has a dew point of 20°C. Its dew point at 30°C and 700 mm Hg will be around __________ °C.

ANS:D - 18.7

To find the dew point at 30°C and 700 mm Hg, we can use the concept of Dalton's Law of Partial Pressures and the Antoine equation for benzene. Given:

• Mixture of 10% C6H6 vapor in air at 25°C and 750 mm Hg has a dew point of 20°C.

First, we need to determine the partial pressure of benzene (𝑃C6H6PC6H6​) in the mixture at 25°C and 750 mm Hg. We can use the given vapor pressure of benzene at 25°C from the Antoine equation: log⁡10(𝑃C6H6)=𝐴−𝐵𝑇+𝐶log10​(PC6H6​)=A−T+CB​ Where:

• 𝐴=6.90565A=6.90565
• 𝐵=1211.033B=1211.033
• 𝐶=220.790C=220.790
• 𝑇T is the temperature in degrees Celsius.

Let's calculate 𝑃C6H6PC6H6​: log⁡10(𝑃C6H6)=6.90565−1211.03325+220.790log10​(PC6H6​)=6.90565−25+220.7901211.033​ log⁡10(𝑃C6H6)=6.90565−1211.033245.790log10​(PC6H6​)=6.90565−245.7901211.033​ log⁡10(𝑃C6H6)=6.90565−4.92403log10​(PC6H6​)=6.90565−4.92403 log⁡10(𝑃C6H6)=1.98162log10​(PC6H6​)=1.98162 𝑃C6H6=101.98162PC6H6​=101.98162 𝑃C6H6=96.658 mm HgPC6H6​=96.658mm Hg Next, we need to find the dew point of the mixture at 30°C and 700 mm Hg. Since benzene obeys Dalton's Law of Partial Pressures, its partial pressure should remain constant when the total pressure decreases to 700 mm Hg. Therefore, the partial pressure of benzene (𝑃C6H6PC6H6​) at 30°C and 700 mm Hg remains 96.658 mm Hg. Now, we can use the Antoine equation again to find the corresponding dew point at 30°C: log⁡10(𝑃C6H6)=6.90565−1211.03330+220.790log10​(PC6H6​)=6.90565−30+220.7901211.033​ log⁡10(𝑃C6H6)=6.90565−1211.033250.790log10​(PC6H6​)=6.90565−250.7901211.033​ log⁡10(𝑃C6H6)=6.90565−4.82798log10​(PC6H6​)=6.90565−4.82798 log⁡10(𝑃C6H6)=2.07767log10​(PC6H6​)=2.07767 𝑃C6H6=102.07767PC6H6​=102.07767 𝑃C6H6=114.749 mm HgPC6H6​=114.749mm Hg Therefore, the dew point at 30°C and 700 mm Hg will be approximately: log⁡10(𝑃C6H6)=6.90565−1211.03330+220.790log10​(PC6H6​)=6.90565−30+220.7901211.033​ log⁡10(𝑃C6H6)=6.90565−1211.033250.790log10​(PC6H6​)=6.90565−250.7901211.033​ log⁡10(𝑃C6H6)=6.90565−4.82798log10​(PC6H6​)=6.90565−4.82798 log⁡10(𝑃C6H6)=2.07767log10​(PC6H6​)=2.07767 𝑃C6H6=102.07767PC6H6​=102.07767 𝑃C6H6=114.749 mm HgPC6H6​=114.749mm Hg Therefore, the dew point at 30°C and 700 mm Hg will be approximately: 𝑇dew=𝐵𝐴−log⁡10(𝑃C6H6)−𝐶Tdew​=A−log10​(PC6H6​)B​−C 𝑇dew=1211.0336.90565−log⁡10(114.749)−220.790Tdew​=6.90565−log10​(114.749)1211.033​−220.790 𝑇dew≈1211.0336.90565−2.06042−220.790Tdew​≈6.90565−2.060421211.033​−220.790 𝑇dew≈1211.0334.84523−220.790Tdew​≈4.845231211.033​−220.790 𝑇dew≈250.09−220.790Tdew​≈250.09−220.790 𝑇dew≈29.3 °CTdew​≈29.3°C Therefore, the dew point at 30°C and 700 mm Hg will be approximately 29.3°C.