What is the vapor pressure of CS2(in mmHg) at 18.5degree C? Carbon disulfide, CS2, has Pvap= 100mm at -5.1 degree C and change in Hvap = 28.0 kj/mol .

Use the Clausius-Clapeyron equation.

To determine the vapor pressure of CS2 at 18.5 degrees Celsius, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)

Where:
- P2 is the vapor pressure at 18.5 degrees Celsius (the unknown)
- P1 is the vapor pressure at -5.1 degrees Celsius (given as 100 mmHg)
- ΔHvap is the heat of vaporization (given as 28.0 kJ/mol)
- R is the ideal gas constant (8.314 J/(mol·K))
- T2 is the temperature in Kelvin at 18.5 degrees Celsius (18.5 + 273.15)
- T1 is the temperature in Kelvin at -5.1 degrees Celsius (-5.1 + 273.15)

First, let's convert the temperatures to Kelvin:
T2 = 18.5 + 273.15 = 291.65 K
T1 = -5.1 + 273.15 = 268.05 K

Now, substitute the given values into the equation:

ln(P2/100) = (-28000 J/mol / 8.314 J/(mol·K)) * (1/291.65 K - 1/268.05 K)

Next, simplify the equation:

ln(P2/100) = (-3364.33) * (0.00343243 - 0.00373228)

Then, calculate the values inside the parentheses:

ln(P2/100) = (-3364.33) * (-0.00029985)

Now, multiply the two values:

ln(P2/100) = 1.00929

To solve for P2, we need to exponentiate both sides to eliminate the natural logarithm:

P2/100 = e^(1.00929)

Finally, solve for P2:

P2 = 100 * e^(1.00929)

Using a calculator, P2 is approximately equal to 151.03 mmHg at 18.5 degrees Celsius.