A certain sample of gas has a volume of 0.452 L measured at 87°C and 0.620 atm. What is its volume at 1 atm and 0°C?

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Answer 1 · 2014-05-19T19:36:38

To solve this problem we would use the combined gas law.

The formula for the combined gas law is
$\frac{(P_{i}) (V_{i})}{T_{i}} = \frac{(P_{f}) (V_{f})}{T_{f}}$ $((P_i)(V_i))/T_i = ((P_f)(V_f))/T_f$

We begin by identifying the values for each of the variables and identifying the missing value.

$P_{i}$ $P_i$ = 0.620 atm
$V_{i}$ $V_i$ = 0.452 L
$T_{i}$ $T_i$ = 87 C + 273 = 360 K

$P_{f}$ $P_f$ = 1 atm
$V_{f}$ $V_f$ = ???
$T_{f}$ $T_f$ = 0 C + 273 = 273 K

$\frac{(0.620 a t m) (0.452 L)}{360 K} = \frac{(1 a t m) (x)}{273 K}$ $((0.620 atm)(0.452 L))/(360 K) = ((1 atm)(x))/(273 K)$

$\frac{(273 K) (0.620 a t m) (0.452 L)}{(360 K) (1 a t m)} = (x L)$ $((273 K)(0.620 atm)(0.452 L))/((360 K)(1atm)) = (x L)$

$x L = 0.212 L$ $x L = 0.212 L$

I hope this was helpful.
SMARTERTEACHER