Q1: At 100°C, water and methylcyclohexane both have vapour pressures of 1 atm. Also at 100°C, the latent heats of vaporisation of these compounds are 40.63 kJ/mole for water and 31.55 kJ/mole for methylcyclohexane. The vapour pressure of water at 150°C is 4.69 atm. At 150°C, the vapour pressure of methylcyclohexane would be expected to be

ANS:C - significantly more than 4.69 atm.

To predict the vapor pressure of methylcyclohexane at 150°C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and the latent heat of vaporization. The equation is as follows: ln⁡(𝑃2𝑃1)=−Δ𝐻vap𝑅(1𝑇2−1𝑇1)ln(P1​P2​​)=−RΔHvap​​(T2​1​−T1​1​) Where:

• 𝑃1P1​ and 𝑃2P2​ are the vapor pressures at temperatures 𝑇1T1​ and 𝑇2T2​ respectively.
• Δ𝐻vapΔHvap​ is the latent heat of vaporization.
• 𝑅R is the gas constant.

Given:

• 𝑃1=1P1​=1 atm (for both water and methylcyclohexane) at 𝑇1=100°𝐶=373.15T1​=100°C=373.15 K
• 𝑃2=4.69P2​=4.69 atm for water at 𝑇2=150°𝐶=423.15T2​=150°C=423.15 K
• Δ𝐻vapΔHvap​ for water = 40.63 kJ/mol
• Δ𝐻vapΔHvap​ for methylcyclohexane = 31.55 kJ/mol

First, we'll calculate the vapor pressure of methylcyclohexane at 150°C using the data given for water: ln⁡(𝑃mch1 atm)=−31.55 kJ/mol8.314 J/mol\cdotpK(1423.15 K−1373.15 K)ln(1 atmPmch​​)=−8.314 J/mol\cdotpK31.55 kJ/mol​(423.15 K1​−373.15 K1​) Solving this equation: ln⁡(𝑃mch1 atm)=−31.55×103 J/mol8.314 J/mol\cdotpK(1423.15 K−1373.15 K)ln(1 atmPmch​​)=−8.314 J/mol\cdotpK31.55×103 J/mol​(423.15 K1​−373.15 K1​) ln⁡(𝑃mch1 atm)≈−3.77ln(1 atmPmch​​)≈−3.77 𝑃mch1 atm≈𝑒−3.771 atmPmch​​≈e−3.77 𝑃mch≈0.0223 atmPmch​≈0.0223 atm Therefore, at 150°C, the vapor pressure of methylcyclohexane would be expected to be significantly less than 4.69 atm. So, the correct answer is significantly less than 4.69 atm.