Respuesta :
Below are supposed to be the question:
a)what is the total pressure of the mixture?
b) calculate the volume in Liters at STP occupied by He and Ne if the N2 is removed selectively.
Below is the answers:
a. Dalton's law says that you just add 'em all together and you will get the total pressures. So it's .89 atm at 15 degrees.
b. Use the partial pressure given and ignore the other gasses. Convert to STP. .32+.42=.74 atm.
This is at 15 degrees C and with a volume of 2.5. We need to change the volume, by the same factor of the gas we just removed. Then we can convert this number to STP. 15/.74=.2. Our gas volume just went down by 20% (by taking out the N2). 2.5-20%=2. The gas is now occupying 2 liters at 15 C and .74 atm.
Let's convert the temperature to K. 15+273=288. When we chill the gas to 0 degrees, we lose some volume, represented by 273/288. So, 2 liters(273/288)=1.9 liters at 0 degrees C and .74 atm.
Now, we need to compress our gas until it reaches 1 atm. This will further reduce the volume. 1.9*(.74/1)=1.4 atm. This is your final answer.
a)what is the total pressure of the mixture?
b) calculate the volume in Liters at STP occupied by He and Ne if the N2 is removed selectively.
Below is the answers:
a. Dalton's law says that you just add 'em all together and you will get the total pressures. So it's .89 atm at 15 degrees.
b. Use the partial pressure given and ignore the other gasses. Convert to STP. .32+.42=.74 atm.
This is at 15 degrees C and with a volume of 2.5. We need to change the volume, by the same factor of the gas we just removed. Then we can convert this number to STP. 15/.74=.2. Our gas volume just went down by 20% (by taking out the N2). 2.5-20%=2. The gas is now occupying 2 liters at 15 C and .74 atm.
Let's convert the temperature to K. 15+273=288. When we chill the gas to 0 degrees, we lose some volume, represented by 273/288. So, 2 liters(273/288)=1.9 liters at 0 degrees C and .74 atm.
Now, we need to compress our gas until it reaches 1 atm. This will further reduce the volume. 1.9*(.74/1)=1.4 atm. This is your final answer.
Answer: The volume occupied at STP is 1.345 L
Explanation:
The equation given by ideal gas follows:
[tex]PV=nRT[/tex] .......(1)
where, P = pressure of the gas
V = Volume of the gas
T = Temperature of the gas
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of gas
- For He gas:
We are given:
[tex]P=0.15atm\\V=2.5L\\T=15^oC=[15+273]=288K\\n=?[/tex]
Putting values in equation 1, we get:
[tex]0.15atm\times 2.5L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 288K\\n=0.016mol[/tex]
- For Ne gas:
We are given:
[tex]P=0.42atm\\V=2.5L\\T=15^oC=[15+273]=288K\\n=?[/tex]
Putting values in equation 1, we get:
[tex]0.42atm\times 2.5L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 288K\\n=0.044mol[/tex]
- Now, calculating the volume occupied by helium and neon at STP when nitrogen gas is removed by using equation 1, we get:
STP conditions:
T = 273 K
P = 1 atm
[tex]n=n_{He}+n_{Ne}=0.016+0.044=0.06mol[/tex]
Putting values in equation 1, we get:
[tex]1atm\times V=0.06mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 273K\\V=1.345L[/tex]
Hence, the volume occupied at STP is 1.345 L
