Final answer:
To find the pressure of an ether system at 300 K using the Clausius-Clapeyron equation, use the known vapor pressure and temperature at the normal boiling point, the heat of vaporization, and the ideal gas constant in the formula. After converting the given heat of vaporization to J/mol, solve for the unknown pressure at the desired temperature.
Step-by-step explanation:
To calculate the pressure of an ether system at 300 K using the Clausius-Clapeyron equation, we must relate the known vapor pressure and temperature at ether's normal boiling point to the desired temperature of 300 K. The equation is:
ln(P2/P1) = -(ΔHvap/R)(1/T2 - 1/T1)
Here, P1 is the pressure at the normal boiling point (760 torr or 101.3 kPa), T1 is the normal boiling point in Kelvin (319 K), P2 is the unknown pressure at 300 K (which we need to find), T2 is the temperature we're interested in (300 K), ΔHvap is the heat of vaporization (given as 29.69 kJ/mol, which needs to be converted to J/mol for using R in J/(mol·K)), and R is the ideal gas constant (8.314 J/(mol·K)). Plugging in all the known values, we can rearrange the equation to solve for P2.
First, convert ΔHvap to J/mol:
ΔHvap = 29.69 kJ/mol × 1000 J/kJ = 29690 J/mol
Next, substitute all the values into the rearranged equation and solve for P2. Remember to convert P1 from kPa to torr by multiplying it by (760 torr / 101.3 kPa) if needed. Finally, using the natural logarithm and algebraic manipulation, calculate P2. The result will be the ether's vapor pressure at 300 K in torr.