The enthalpy of vaporization of substance X is 5.00 kJ per mole and its normal boiling point is -57°C. Calculate the vapor pressure of X at -79°C round your answer to 2 significant digits
To solve this problem, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = enthalpy of vaporization
R = ideal gas constant (8.314 J/mol·K)
First, we need to convert the temperatures to Kelvin:
T1 = -57°C = 216 K
T2 = -79°C = 194 K
Now, we can substitute the values into the equation:
ln(P2/P1) = (5.00 × 10^3 J/mol) / (8.314 J/mol·K) * (1/216 K - 1/194 K)
ln(P2/P1) = (5.00 × 10^3 J/mol) / (8.314 J/mol·K) * (0.004629 - 0.005155)
ln(P2/P1) = (5.00 × 10^3 J/mol) / (8.314 J/mol·K) * (-0.000526)
ln(P2/P1) = -3.161
Now, we can solve for P2/P1 by taking the exponential of both sides:
P2/P1 = e^(-3.161)
P2/P1 = 0.042
To find P2, we can assume P1 = 1 atm, so P2 = 0.042 atm.
Therefore, the vapor pressure of substance X at -79°C is approximately 0.042 atm.