Question

3. A sample of krypton gas occupies 75.0 mL at 0.400 atm. If the temperature remained constant, what volume would the krypton occupy at:

c. 765 mmHg; d. 4.00 mmHg; e. 3.50x10-2

Answer 1

c) 29.8 mL

d) 5375 mL

e) $6.5\cdot 10^5 mL$

Explanation:

c)

We can solve this problem by using Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

$pV=const.$

where

p is the pressure of the gas

V is its volume

We can rewrite the formula as

$p_1 V_1 = p_2 V_2$

For the gas in this problem:

$p_1=0.400 atm$ is the initial pressure

$V_1=75.0 mL$ is the initial volume

$p_2=765 mmHg = 1.006 atm$ is the final pressure (using the conversion factor $1 atm = 760 atm$)

Solving for V2, we find the final volume:

$V_2=\frac{p_1 V_1}{p_2}=\frac{(0.400)(75.0)}{1.006}=29.8 mL$

d)

We can solve this part by using again the equation:

$p_1 V_1 = p_2 V_2$

Where in this case we have:

$p_1=0.400 atm$ is the initial pressure

$V_1=75.0 mL$ is the initial volume

$p_2=4.00 mmHg$ is the final pressure

Converting into atmospheres,

$p_2 = 4.00 mmHg \cdot \frac{1}{760 mmHg/atm}=0.0053 atm$

And solving for V2, we find the final volume:

$V_2=\frac{p_1 V_1}{p_2}=\frac{(0.400)(75.0)}{0.0056}=5357 mL$

e)

As before, we use Boyles' Law:

$p_1 V_1 = p_2 V_2$

In this part we have:

$p_1=0.400 atm$ is the initial pressure of the gas

$V_1=75.0 mL$ is the initial volume of the gas

$p_2=3.50\cdot 10^{-2} torr$

1 torr is equivalent to 1 mmHg, so the conversion factor is the same as before, therefore the final pressure in atmospheres is:

$p_2 = 3.50\cdot 10^{-2} mmHg \cdot \frac{1}{760 mmHg/atm}=4.6\cdot 10^{-5} atm$

And so, the final volume of the krypton gas is:

$V_2=\frac{p_1 V_1}{p_2}=\frac{(0.400)(75.0)}{4.6\cdot 10^{-5}}=6.5\cdot 10^5 mL$