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

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.