In an open system the vapor pressure is equal to the

Chemistry

1 answer:

erica [24]3 years ago

5 0

Answer: In an open system the vapor pressure is equal to the outside air pressure.

Explanation:

An open system is a system that is capable of exchanging both matter and energy with its surroundings.

For example, a hot cup of coffee is an open system as the cup is not covered with a lid. Therefore, heat energy from the coffee goes into the surroundings.

Vapor pressure is defined as the ability of a substance to change into vapor state. pressure

Therefore, we can conclude that in an open system the vapor pressure is equal to the outside air pressure.

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An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a

Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):

$V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol$

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

$n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol$

$V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21$