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

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?

1 Answer

Impact of this question

8544 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-24T02:57:11

Its volume at STP is 28.09L.

Explanation:

We first need to find the volume of the $C O_{2}$ $CO_2$ sample at the initial conditions.

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

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

We also need the moles of gas, $n$ $n$ . To do this, we need the molar mass of $C O_{2}$ $CO_2$ , which is 44.01 g/mol.
$n = \frac{m}{M}$ $n = m/M$
$n = \frac{44.0 g}{44.01 \frac{g}{m o l}}$ $n = (44.0g) / (44.01g/(mol))$
$n = 1.00 m o l$ $n = 1.00mol$

The ideal gas constant, $R$ $R$ , is equal to $8.314 \frac{L \cdot k P a}{m o l \cdot K}$ $8.314(L * kPa)/(mol * K)$ .

Substitute these values into the ideal gas law equation, $P V = n R T$ $PV = nRT$ :

$(86.0 k P a) V = (1.00 m o l) (8.314 \frac{L \cdot k P a}{m o l \cdot K}) (338 K)$ $(86.0kPa)V = (1.00mol)(8.314(L*kPa)/(mol*K))(338K)$ .

Rearrange for V:
$V = \frac{(1.00 m o l) (8.314 \frac{L \cdot k P a}{m o l \cdot K}) (338 K)}{86.0 k P a}$ $V = ((1.00mol)(8.314(L*kPa)/(mol*K))(338K))/(86.0kPa)$
$V = 32.66 L$ $V = 32.66L$ .

We can call this the initial volume, $V_{i}$ $V_i$ .

We can now use any of the gas law equations to solve the problem. Let's use Boyle's Law:
$P_{i} V_{i} = P_{f} V_{f}$ $P_iV_i = P_fV_f$ .

We can substitute in $V_{i} = 32.66 L$ $V_i = 32.66L$ , $P_{i} = 86.0 k P a$ $P_i = 86.0kPa$ . At STP (standard temperature and pressure), $P = 100.0 k P a$ $P = 100.0kPa$ , so we also substitute $P_{f} = 100.0 k P a$ $P_f = 100.0kPa$ .
$(86.0 k P a) (32.66 L) = (100.0 k P a) V_{f}$ $(86.0kPa)(32.66L) = (100.0kPa)V_f$

Rearrange for $V_{f}$ $V_f$ :
$V_{f} = \frac{(86.0 k P a) (32.66 L)}{100.0 k P a}$ $V_f = ((86.0kPa)(32.66L))/(100.0kPa)$
$V_{f} = 28.09 L$ $V_f = 28.09L$ .

The final volume is 28.09L.

Science

Math

Humanities

... and beyond

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

Explanation:

Related questions

Impact of this question