Question

Consider an ideal gas that exerts a pressure of \(23.76 \mathrm{~mm}\) \(\mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\). Assuming \(n\) and \(V\) are held constant, what would its pressure be at \(40^{\circ} \mathrm{C} ? 70^{\circ} \mathrm{C} ? 100^{\circ} \mathrm{C} ?\) Compare the numbers you have just calculated with the vapor pressures of water at these temperatures. Can you suggest a reason why the two sets of numbers are so different?

Short answer

Answer: The vapor pressures of water differ from the pressures of an ideal gas at the same temperatures because water molecules have strong intermolecular forces, such as hydrogen bonding, which cause the actual behavior of water to deviate from the ideal gas behavior. In contrast, the ideal gas law assumes there are no interactions between the gas molecules, thus leading to different pressure values.

Step-by-step solution

Understand the ideal gas law.

The ideal gas law is given by the formula: PV = nRT Here, P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. In this exercise, we are told that n and V are held constant. This means any changes in pressure are directly related to changes in temperature.

Express pressure in terms of temperature.

Since n, V, and R are constants, we can write the ratio of pressures and temperatures as: P1/T1 = P2/T2 where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature. We will use this formula to find the pressure of the gas at different temperatures.

Convert the given temperature and pressure to Kelvin and atm.

The pressures provided are in mmHg, and temperatures are in Celsius. We need to convert them to standard units: atm and Kelvin. 1 atm = 760 mmHg, so 23.76 mmHg * (1 atm / 760 mmHg) = 0.03126 atm. To convert Celsius to Kelvin, add 273.15 to the Celsius temperature: T1 = 25°C + 273.15 = 298.15 K Now we have P1 = 0.03126 atm and T1 = 298.15 K.

Calculate the pressure at each temperature.

First, let's calculate the pressure at each temperature required: - T2_40 = 40°C + 273.15 = 313.15 K - T2_70 = 70°C + 273.15 = 343.15 K - T2_100 = 100°C + 273.15 = 373.15 K Using P1/T1 = P2/T2, and solving for P2, we get: P2 = P1 * T2/T1 For each temperature: - P2_40 = 0.03126 atm * 313.15 K / 298.15 K = 0.03267 atm - P2_70 = 0.03126 atm * 343.15 K / 298.15 K = 0.03409 atm - P2_100 = 0.03126 atm * 373.15 K / 298.15 K = 0.03550 atm Now convert back to mmHg: - P2_40 = 0.03267 atm * 760 mmHg/atm = 24.83 mmHg - P2_70 = 0.03409 atm * 760 mmHg/atm = 25.91 mmHg - P2_100 = 0.03550 atm * 760 mmHg/atm = 26.98 mmHg

Compare with vapor pressures of water.

Now we can compare these pressures with vapor pressures of water. The vapor pressures of water at the given temperatures are: - 40°C: ~55 mmHg - 70°C: ~233 mmHg - 100°C: 760 mmHg Comparing these values with our calculated pressures, we can see that the vapor pressures of water are significantly higher than the pressures of the ideal gas at the same temperatures. The reason for this difference is that the ideal gas law assumes there are no interactions between the gas molecules, while water molecules have strong intermolecular forces (hydrogen bonding). This makes the actual behavior of water deviate from the ideal gas behavior, resulting in higher vapor pressures observed for water compared to ideal gases at the same temperature.