Final answer:
The limiting value of the ratio of the pressure of the gas at the boiling point of water to its pressure at the triple point of water can be determined using the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance to its temperature and enthalpy of vaporization. By rearranging the equation and plugging in the given temperatures, we can calculate the desired ratio.
Step-by-step explanation:
The limiting value of the ratio of the pressure of the gas at the boiling point of water to its pressure at the triple point of water can be determined using the Clausius-Clapeyron equation. The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature and enthalpy of vaporization. In this case, we are given the temperature at the boiling point of water (373.15 K) and the triple point of water (273.16 K), and we need to find the ratio of the pressures.
Using the Clausius-Clapeyron equation, we have:
ln(P₂/P₁) = (ΔHvap/R) * (1/T₁ - 1/T₂)
Where P₁ and P₂ are the pressures at the given temperatures, T₁ and T₂ are the temperatures in Kelvin, ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. We can assume that the volume of the gas is the same at both temperatures.
Since we know the temperature at the boiling point of water (373.15 K) and want to find the temperature at the triple point of water (273.16 K), we can rearrange the equation and solve for P₂ / P₁:
P₂ / P₁ = exp((ΔHvap/R) * (1/T₁ - 1/T₂))