Question

A 14.0 L sample of fluorine gas has a pressure of 29.0 atm at a certain temperature. At the same temperature, what volume would this gas occupy at a pressure of 2.90 atm? Assume ideal behavior. V=

Answer 1

Answer:

The volume occupied by the gas at a pressure of 2.90 atm = 140L

Explanation:

Since we are assuming ideal behavior, the gas equation is:

pV = nRT

where, p is the pressure

           V is the volume

           R is the universal gas constant

            T is the temperature

The temperature is constant. We can use the Boyle's Law:

$p_{1}V_{1}=p_{2}V_{2}$

$V_{2}=\frac{p_{1}V_{1} }{p_{2}}$ -----(1)

$p_{1}=29.0$ atm

$V_{1}=14L$

$p_{2}=2.90$ atm

$V_{2}=\frac{29.0*14}{2.90}$=$140L$

Therefore, at the same temperature, the volume occupied by the gas at a pressure of 2.90 atm is 140L.