Use the Clausius-Clapeyron equation to cal-

culate the temperature at which pure water
would boil at a pressure of 434 torr. The
heat of vaporization of water is 41276 J/mol
at 100�C.
Answer in units of �C

What's your problem with this? You just plug and chug.

p1 = 760 with T1 = 373 K
p2 = 434 with T2 = ? K

DrBob222 needs to chill

Doctor Bob a mf savage

Stupid DrBob222, this isn't a place where students come to check out how to do it, most people just want the answer, so shut up and tell us how to do it but also give the answer

To calculate the temperature at which pure water would boil at a given pressure of 434 torr using the Clausius-Clapeyron equation, we need to use the following formula:

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

Where:
P2 = the given pressure (434 torr)
P1 = the standard pressure (which is typically 1 atm or 760 torr)
ΔHvap = heat of vaporization of water (41276 J/mol)
R = gas constant (8.314 J/(mol*K))
T2 = the boiling temperature we want to calculate
T1 = the standard boiling temperature of water (100�C or 373 K)

First, we need to convert the given pressure to atm, as the gas constant R is typically expressed in atm units. Since 1 atm is equal to 760 torr, we have:

P2 = 434 torr / 760 torr/atm
P2 = 0.5705 atm

Now we can plug the values into the equation:

ln(0.5705 atm / 1 atm) = (-41276 J/mol / (8.314 J/(mol*K))) * (1/T2 - 1/373 K)

Simplifying:

ln(0.5705) = (-41276 / 8.314) * (1/T2 - 1/373)

Now, rearrange the equation to solve for 1/T2:

1/T2 = (ln(0.5705) * 8.314 / -41276) + 1/373

1/T2 = -0.002181 + 0.002680

1/T2 = 0.000499

Finally, rearrange the equation again to calculate T2:

T2 = 1 / (0.000499)
T2 ≈ 2004 K

To convert from Kelvin to Celsius, subtract 273 from the result:

T2 ≈ 1731 �C

Therefore, the temperature at which pure water would boil at a pressure of 434 torr is approximately 1731 �C.