Question

missed this? read section 6.6 (page), watch kcv 6.6 a 280 - ml flask contains pure helium at a pressure of 762 torr. a second flask with a volume of 490 ml contains pure argon at a pressure of 712 torr. if we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of helium? express the partial pressure in torr to three significant figures. part b if we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of argon? express the partial pressure in torr to three significant figures.

Explanation:

Step1: Use Boyle's Law ($P_1V_1 = P_2V_2$) for helium

For helium, $P_1 = 762$ torr, $V_1=280$ mL, and $V_2=(280 + 490)$ mL.
We can solve for $P_{He}$ (the new partial pressure of helium) using the formula $P_{He}=\frac{P_1V_1}{V_2}$.
Substitute the values: $P_{He}=\frac{762\times280}{280 + 490}$.

Step2: Calculate the value for helium

First, calculate the denominator: $280+490 = 770$ mL.
Then, $762\times280=213360$.
So, $P_{He}=\frac{213360}{770}\approx277$ torr.

Step3: Use Boyle's Law ($P_1V_1 = P_2V_2$) for argon

For argon, $P_1 = 712$ torr, $V_1 = 490$ mL, and $V_2=(280 + 490)$ mL.
We can solve for $P_{Ar}$ (the new partial pressure of argon) using the formula $P_{Ar}=\frac{P_1V_1}{V_2}$.
Substitute the values: $P_{Ar}=\frac{712\times490}{280 + 490}$.

Step4: Calculate the value for argon

First, calculate $712\times490 = 348880$.
Then, $P_{Ar}=\frac{348880}{770}\approx453$ torr.

Answer:

$P_{He}=277$ torr
$P_{Ar}=453$ torr