A 2.0 liter bottle can withstand a pressure of 5.0 atm. 120 mL of liquid ethanol, #"C"_2"H"_5"OH"#, (density 0.789 g/mL), is poured into the bottle and sealed. The bottle is heated to 100 C, changing all the liquid to gas. Will the bottle explode?

#120 cancel("mL C"_2"H"_5"OH")*(0.789cancel( "g C"_2"H"_5"OH"))/(cancel("mL C"_2"H"_5"OH"))*("mol C"_2"H"_5"OH")/(46.07cancel("g C"_2"H"_5"OH")) = 2.055 "mol C"_2"H"_5"OH" #

Alright, now we will use the Ideal gas law and solve for Pressure,

#P=(nRT)/V#

#(2.055 cancel("mol C"_2"H"_5"OH")*(0.082cancel(L)"atm")/(cancel("mol")cancel("K"))*373cancel( K))/ (2cancel(L))#= #31.4 "atm"#

Yes, the bottle will burst, in the calculation, 373 K was used because that's the temperature at which ethanol was all gas, and so, therefore, the gas law would be applied and then once it was a gas, it automatically occupied the volume of the bottle which was 2 L.

(@Hi's answer was awesome—this answer just elaborates more on why they used the values and equation that they did.)

This is what we're given:

• #V#, or volume, of #"2.0 L"#. We consider this to be the volume of our gas, because a property of the gas phase is that its volume freely expands to fill up the container.
  In our case, our container was #"2.0 L"#, so the volume of gas will also be #"2.0 L"#
• #T#, or temperature, which is #100°C# or #100 + 273.15 = "373.15 K"#.
• The volume of #C_2H_5OH# as a liquid: #"120 mL"#.
• The density of #C_2H_5OH# as a liquid: #"0.789 g/mL"#.
• #R#, or the Ideal Gas Constant, which is #"0.08206 L atm/g mol"#. (Quick note: We always know the value of #R# because it's a constant.)

To determine if the bottle will explode, we need to find the pressure of gaseous #C_2H_5OH#.
If it exceeds #"5.0 atm"#, the bottle will explode. If it doesn't, the bottle won't explode.

We can use the Ideal Gas Law to find the pressure of #C_2H_5OH#:

#pV = nRT#

And rearrange this to have only #p#, pressure, on one side:

#p = (nRT)/V#

We already know #R#, #T# #V#, but a bit more work needs to be done to find the value of #n#, or the number of moles of gas.

To find moles, we first need to find mass.
We can use the density equation:

#"density" = "mass"/"volume"#

Therefore, #"mass" = "volume" xx "density"#.

Since density and volume were given in the question:

#"mass" = "120 mL" xx "0.789 g/mL" = 94.68 g"#

Then, we can use mass to find #n#:

#n = "mass"/"molar mass"#
#n = ("94.68 g")/((Cxx2+Hxx5+O+H)"g/mol")#
#n = ("94.68 g")/((12.01xx2+1.008xx5+16.00+1.008)"g/mol")#
#n = "2.055 mol"#

Finally, we can plug #V#, #n#, #R#, and #T#, into the Ideal Gas Law:

#p = ("2.055 mol" xx "0.08206 L atm/K mol" xx "373.15 K")/"2.0 L"#

#p = (2.055 cancel("mol") xx 0.08206 cancel("L") "atm/"cancel("K mol") xx 373.15 cancel("K"))/(2.0 cancel("L"))#

#p = "31.47 atm"#

Since the pressure exceeds #"5.0 atm"# (by a lot!), the bottle will (definitely) explode.