A carbon dioxide sample weighing 44.0 g occupies 32.68 L at 65°C and 645 torr. What is its volume at STP?

1 Answer
Jun 24, 2018

Answer:

Its volume at STP is 28.09L.

Explanation:

We first need to find the volume of the #CO_2# sample at the initial conditions.

To do this, we can use the ideal gas law equation, #PV = nRT#, where #P# = pressure, #V# = volume, #n# = moles of gas, #R# is the ideal gas constant, and #T# = temperature in Kelvins.

We should first convert our units to standard gas units.
#P# = 645 Torr = 86.0 kPa.
#T# = 65 C = 338K.

We also need the moles of gas, #n#. To do this, we need the molar mass of #CO_2#, which is 44.01 g/mol.
#n = m/M#
#n = (44.0g) / (44.01g/(mol))#
#n = 1.00mol#

The ideal gas constant, #R#, is equal to #8.314(L * kPa)/(mol * K)#.

Substitute these values into the ideal gas law equation, #PV = nRT#:

#(86.0kPa)V = (1.00mol)(8.314(L*kPa)/(mol*K))(338K)#.

Rearrange for V:
#V = ((1.00mol)(8.314(L*kPa)/(mol*K))(338K))/(86.0kPa)#
#V = 32.66L#.

We can call this the initial volume, #V_i#.

We can now use any of the gas law equations to solve the problem. Let's use Boyle's Law:
#P_iV_i = P_fV_f#.

We can substitute in #V_i = 32.66L#, #P_i = 86.0kPa#. At STP (standard temperature and pressure), #P = 100.0kPa#, so we also substitute #P_f = 100.0kPa#.
#(86.0kPa)(32.66L) = (100.0kPa)V_f#

Rearrange for #V_f#:
#V_f = ((86.0kPa)(32.66L))/(100.0kPa)#
#V_f = 28.09L#.

The final volume is 28.09L.