The Clapeyron-Clausisus Equation is ln(P2/P1) = dHvap/R [(1/T1)-(1/T2)]. Assuming that dHvap doesn't change with the temperature, this equation relates the change in vapor pressure and temperature to a substances enthalpy of vaporization (R is the molar gas constant). Use this relationship and the fact that dHvap of water at 25C is 43.99 kL/mole to calculate the vapor pressure of water at 5, 25, 50, 95 C. Results should be in order from least to greatest and in Torr.
(A) 21.2, 39.6, 86, 731
(B) 5.92, 39.6, 92, 601
(C) 5.92, 21.2, 83.9, 622
(D) none of these
The one piece of data you don't have in the problem (no one ever puts this in and that's the secret) is you are supposed to recall that the boiling point occurs when the vapor pressure of a liquid equals atmospheric pressure. The piece of data you need is pH2O @ 100 C = 760 torr.
Substitute this and the other data and solve.
I don't know how to solve it
Plug in the numbers and post it.
plug what into where? my teacher hasn't explained anything relating to this problem so I have no idea what to do.
To calculate the vapor pressure of water at different temperatures using the Clapeyron-Clausius equation, we need to use the provided enthalpy of vaporization (ΔHvap) of water at 25°C, which is 43.99 kJ/mol.
First, let's convert the enthalpy of vaporization from kilojoules per mole (kJ/mol) to joules per mole (J/mol) by multiplying by 1000:
ΔHvap = 43.99 kJ/mol × 1000 J/kJ = 43,990 J/mol
Next, we need to convert ΔHvap to liters per mole (L/mol) by dividing by the molar volume of water, which is approximately 18 L/mol at standard temperature and pressure:
ΔHvap = 43,990 J/mol ÷ 18 L/mol = 2,443.9 J/L ≈ 2.44 kJ/L
Now we can use the Clapeyron-Clausius equation to calculate the vapor pressure (P2) at different temperatures (T2) by setting the initial pressure (P1) to 1 atm and the initial temperature (T1) to 25°C (298 K). We will use the gas constant (R) which is approximately 0.0821 L·atm/(mol·K).
Let's calculate the vapor pressure at different temperatures:
1. At 5°C (278 K):
ln(P2/1 atm) = (2.44 kJ/L)/(0.0821 L·atm/(mol·K)) × [(1/298 K) - (1/278 K)]
ln(P2) = 30.4 mol J/(mol·K) × [(1/298 K) - (1/278 K)]
ln(P2) ≈ 30.4 mol J/(mol·K) × (0.00336 K^(-1))
ln(P2) ≈ 0.102
P2 ≈ e^(0.102) ≈ 1.107 atm ≈ 842.6 Torr
2. At 25°C (298 K):
ln(P2/1 atm) = (2.44 kJ/L)/(0.0821 L·atm/(mol·K)) × [(1/298 K) - (1/298 K)]
ln(P2) = 30.4 mol J/(mol·K) × (0)
ln(P2) = 0
P2 = e^0 = 1 atm = 760 Torr
3. At 50°C (323 K):
ln(P2/1 atm) = (2.44 kJ/L)/(0.0821 L·atm/(mol·K)) × [(1/298 K) - (1/323 K)]
ln(P2) ≈ 30.4 mol J/(mol·K) × (-0.00380 K^(-1))
ln(P2) ≈ -0.116
P2 ≈ e^(-0.116) ≈ 0.891 atm ≈ 676.4 Torr
4. At 95°C (368 K):
ln(P2/1 atm) = (2.44 kJ/L)/(0.0821 L·atm/(mol·K)) × [(1/298 K) - (1/368 K)]
ln(P2) ≈ 30.4 mol J/(mol·K) × (-0.00838 K^(-1))
ln(P2) ≈ -0.255
P2 ≈ e^(-0.255) ≈ 0.774 atm ≈ 587.8 Torr
Now that we have the vapor pressures at 5°C, 25°C, 50°C, and 95°C, let's compare them with the given answer choices:
(A) 21.2, 39.6, 86, 731
(B) 5.92, 39.6, 92, 601
(C) 5.92, 21.2, 83.9, 622
(D) none of these
From our calculations, we can see that the closest answer choice is (C) 5.92, 21.2, 83.9, 622.