Q1: The vapour pressure of water is given by, in P_sat = A - (5000/T), where A is a constant, Psat is the vapour pressure in atm. and T is the temperature in K.The vapor pressure of water in atm. at 50°C is approximately

ANS:D - 0.13

To find the vapor pressure of water at 50°C using the given equation 𝑃sat=𝐴−5000𝑇Psat​=A−T5000​, we need to first convert 50°C to Kelvin. 𝑇=50°𝐶+273.15=323.15 KT=50°C+273.15=323.15K Now, we can plug this value of 𝑇T into the equation and solve for 𝑃satPsat​. Given that 𝐴A is a constant and not provided, we cannot directly calculate the exact value of 𝑃satPsat​. However, we can make an approximation using the given options. 𝑃sat≈𝐴−5000323.15Psat​≈A−323.155000​ We'll check which option is closest to the result we obtain. Let's calculate 5000323.15323.155000​: 5000323.15≈15.46323.155000​≈15.46 We subtract this from the given options to see which one is closest: (a) 0.07−15.460.07−15.46 - Not possible as the result is negative. (b) 0.09−15.460.09−15.46 - Not possible as the result is negative. (c) 0.11−15.46=−15.350.11−15.46=−15.35 - Still negative. (d) 0.13−15.46=−15.330.13−15.46=−15.33 - Negative. Based on these calculations, none of the given options result in a positive value, suggesting there may be a mistake in the provided equation or the options. Please double-check the equation or provide additional information if available.