Question

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

Answer 1

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.