A sealed flask contains 0.60 g of water at 28 degrees celcius. The vapor pressure of water at this temperature is 28.36 mmHg.
What is the minimum volume of the flask in order that no liquid water be present in the flask?
2 answers
PV=nRT should be used
v=nrt/p
v= (0.03333)(0.0821)(301)/0.037316
v= 22.07
n- 0.6 divided by the molar mass of water (18) to get moles
r- constant (0.0821)
t- temperature in kelvin ( 28 + 273)
p- pressure in atm ( convert mm mercury to atm by dividing 28.36 by 760)
v= (0.03333)(0.0821)(301)/0.037316
v= 22.07
n- 0.6 divided by the molar mass of water (18) to get moles
r- constant (0.0821)
t- temperature in kelvin ( 28 + 273)
p- pressure in atm ( convert mm mercury to atm by dividing 28.36 by 760)
0 answers