Calculate the amount of heat needed to transform, under standard conditions, 1.0 mol of liquid methanol at 273 K to 1.0 mol of methanol vapor at 373 K. Assume that Cp,m for liquid methanol and for methanol vapor are independent of temperature. (Enthalpy of vaporization of methanol 35.3 KJ/mol-1, Cp,m of methanol 81.6 JK-1/mol)
q1 = no of moles*Cp*
T
= 1*81.6*(373-273)
= 8160joules = 8.61KJ
q2 = no of moles*Enthalpy of vaporization of methanol
= 1*35.3 KJ/mol = 35.3KJ
q = q1 +q2
= 8.61 + 35.3
= 43.91Kj >>>>> answer
Calculate the amount of heat needed to transform, under standard conditions, 1.0 mol of liquid methanol...
The Standard enthalpy of vaporization of water at 100.0
oC is 40.66 KJ*mol-1. The Cp,m
values for the liquid and the vapor water are, respectively, 75.3
and 33.58 J*K-1*mol-1. Assume that the heat
capacities are independent of temperature, and that the vapor
behaves as an ideal gas.
a) Calculate
sys in taking one mole of liquid water at 25.0
oC and 1.00 atm to gaseous water at 95.0 oC
and 0.500 atm.
b) Assume that the temperature and pressure of...
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the
vapor pressure of acetonitrile changes with temperature at a rate
of 0.03 atm / K near its normal boiling point of 80C. Under these
conditions the heat of vaporization would be?
a) 31 KJ/mol b) 13 KJ/mol c) 41 KJ/mol d) 0.03 KJ/mol
a) 31 KJ/mol b) 13 KJ/mol c) 41 KJ/mol d) 0.03 KJ/mol
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