Mixture of hydrogen and oxygen in container

[Numeriprimi]

Homework Statement

In a closed container with a volume of V = [STRIKE]5[/STRIKE] 50 l is a mixture of hydrogen and oxygen with the total mass m = 50 g. The container have pressure p = 300 000 Pa and the temperature t_1 = 20 ° C.
A) What is the number of moles n_1 (hydrogen) and n_2 (oxygen) in the container?
B) What heat Q is released after an electrical spark ignited the mixture? The calorific value of the hydrogen is 120 000 000 J / kg.
C) What is the pressure in the container when the temperature go down to t_2 = 100 ° C (after elektrical spark) and what is m_v weight of condensed water at this temperature?

... p_a = 1, 013 * 100 000 Pa; M_mhydrohen = 2,02/1000 kg/mol; M_moxygen = 32/1000 kg/mol; R = 8,31 J/(mol*K)

Homework Equations

A) pV = nRT
B) Q = Hm
C) p/T = const.

The Attempt at a Solution

A) pV = nRT = m/M_m * RT; m/V = pM_m/ RT; m/V(hydrogen) = pM_mhydrogen/RT; V_(oxygen) = pM_moxygen/RT
The system of equations:
- V_hydrogen +V_oxygen = V (V_oxygen = V - V_hydrogen ... substituted into the second equation)
- V_hydrogen*(p*M_mhydrogen/RT) + V_oxygen*(p*M_moxygen/RT) = m
Ok, I solve the system of equation... And I get V_oxygen = 0,01 m^3; V_hydrogen = 0, 04 m^3
Now back to equation pV = nRT; n = pV/2RT (2 because it H_2 and O_2, it is right?)
And I get n_oxygen = 0, 626 mol; n_hydrogen = 2, 45 mol
It is OK?
B) Q = Hm = H*n*M_mhydrogen*2 (2 because it is H_2, it is right?)
Q = 1 187 760 J... It is OK?
C) Hmmm... And this is the biggest problem.
It is isochoric progress? Than p_1/T_1 = p_2/T_2; p_2 = p_1*T_2/T_1
p_2 = 300 000 * (273,15 + 100)/(273, 15 + 20) Pa = 381 867 Pa
Condensed water... Hmm, basic assumption is that the system is in steady state - System of saturated steam and liquid. Saturated steam is not an ideal gas ... The equation p / T = const. ist probably wrong. I really don't know any equation for saturated steam... Can you help me?
Thank you and sorry for my bad English :-)

 

[SteamKing]

How many liters are in 1 m^3?

 

[Numeriprimi]

1 l = 1 dm^3, so 1000 l = 1 m^3...
yeees, error, 5 l = 0, 005 m^3... Thank you.
However, the process, is good? And what with C?

 

[Chestermiller]

Maybe I'm doing something wrong, but I calculated that the total number of moles of H2 plus O2 in the container is 0.616 gm-moles. However, if x is the number of grams of H2, and 50-x is the number of grams of O2, there was no combination that gives 0.616 moles. Are you sure about that 50 grams? Are you sure it's not 5 gm?

Chet

 

[Numeriprimi]

Mistake, big mistake in task...
In a closed container with a volume of V = 50 l ... Other info is OK.
So, what now, it is good?

 

[Chestermiller]

In that case, I get 1.26 moles of O2 and 4.9 moles of H2.

 

[Numeriprimi]

Hmmm, so, you get twice larger values (my are 0, 626 moles and 2,45 moles). I divide n by 2, because I consider consider H2 and O2. The resoults 1, 26 moles and 4,9 moles are only for H and O, I think? Or no?

 

[Numeriprimi]

And what about part B and C? Do you have any idea?

 

[Chestermiller]

Hmmm, so, you get twice larger values (my are 0, 626 moles and 2,45 moles). I divide n by 2, because I consider consider H2 and O2. The resoults 1, 26 moles and 4,9 moles are only for H and O, I think? Or no?

No. From the ideal gas law, the total number of moles is 6.16, irrespective of the individual molecular weights: (300000)(0.05)/(8.31)(293). So figure out where you went wrong.

 

[Numeriprimi]

Ok, ok... The mistake was the n dividet two, now I have same resoults.

 

[Chestermiller]

There doesn't seem to me to be enough data provided to properly solve part B. But, I can help you with part C.
After the reaction is complete, how many moles each of H2, O2, and H2O are in the container? If there is liquid water present in the container at the final temperature of 100 C, what is the partial pressure of the water vapor in the gas phase?

Please answer these questions first, and then we can continue.

Chet

 

[Numeriprimi]

Ok, so I have some solution...
First - how many moles has each of H2, O2, and H2O?
m=m, (total mass of container in part A = total mass of container in part C)
I have to exprees m,... It is sum of m_H, + m_O, + m_H2O, in part C.
1) hydrogen: m_H = n * M_H = N,*M_H/N_A ... However, we can write N,=N-2N,, (N, is number of partiples H2 in part c, N is nimber of partiples H2 in part a; 2N, is number of partiples for water)
m_H = (N-2N,,)*M_H/N_A= (n_H - 2N,,/N_A)*M_H = n_H*M_H -2N,,*M_H/N_A (n_H is number of moles of hydrogen from part a)
2) oxygen: m_O = (N-N,,)*M_O/N_A = (n_O - N,,/N_A)*M_O = n_O*M_O -N,,*M_O/N_A ( n_H is number of moles of oxyhen from part a; ... 2N,, of H_2 and N,, of O_2 for H2O)
3) water: m_H2O = 3N,,*M_H20/N_A

m = m_H + m_O + m_H2O = (n_H*M_H -2N,,*M_H/N_A)+(n_O*M_O -N,,*M_O/N_A)+(3N,,*M_H20/N_A)
m - n_H*M_H - n_O*M_O = -2N,,*M_H/N_A - N,,*M_O/N_A + 3N,,*M_H2O/N_A
N_A(m - n_H*M_H - n_O*M_O) = N,,(-2M_H - M_O + 3M_H2O)
N,, = N_A(m - n_H*M_H - n_O*M_O)/(-2M_H - M_O + 3M_H2O)
N,, = 6,022*10^23(0,05 - 4,9*2,02*10^-3 - 1,26*32*10^-3)/(-2*2,02*10^-3 - 32*10^-3 + 3(2*2,02*10^-3+32*10^-3) = 6,022*10^23 * -2,18*10^-4 / 0,07208 = -1,82*10^21 is the number of moles in hydrogen, hydrogen and oxygen we can easily calculate, but ... Eeeeemh. Crazy resoult. Hmmm... Something is wrong, don't know what.
If i use my twice smaller values: n_oxygen = 0, 626 mol; n_hydrogen = 2, 45 mol
Water: 2, 1*10^23 ... 0, 35 moles
Oxygen: (0,626-0,35)moles = 0,276 moles
Hydrogen: (2,45-2*0,35)moles = 1,75 moles
... Hmmm. I don't understand why, but i think the idea is good. May I am too sleepy and thus stupit :-D
Do you have some resoult? Is my idea good or do you see a mistake? Thank you and good night.

 

[Chestermiller]

What you did is a lot of work, but I can't make sense out of it because of your complicated notation. Anyway, here is my simple reasoning: To start with, you have 1.26 moles of O2 and 4.9 moles of hydrogen. The total number of moles does not stay constant since 2 moles of H2 reacts with 1 mole of O2 to produce 2 moles of H20. The limiting reactant is oxygen, so it all gets used up. This consumes (1.26)(2)=2.52 moles of H2 to produce 2.52 moles of H2O. So, in the final state, you have 0 moles of O2, 4.9 - 2.52 = 2.38 moles of H2, and 2.52 moles of H20. The final number of moles in the container is 4.9.

Next, please answer my question about the partial pressure of the water vapor in the final state of the system.

 

[Numeriprimi]

Eeeeh. So simly? Thank you. I will answer your next question for a few hours, I go sleep for a while... However, why do you think that all oxygen have to be consumed? You don't know, when on the container is equilibrium state. Equilibrium state can be before consumed all oxygen, no?

 

[Chestermiller]

Eeeeh. So simly? Thank you. I will answer your next question for a few hours, I go sleep for a while... However, why do you think that all oxygen have to be consumed? You don't know, when on the container is equilibrium state. Equilibrium state can be before consumed all oxygen, no?

This reaction is well-known to go virtually to completion. It has an extremely high equilibrium constant. So the amount of O2 remaining is going to be insignificant.

Chet

 

[Numeriprimi]

The partial pressure of water vapor = pressure of water vapor in the V of the system and with temperature T system... Is it right? However, why particial pressure? It isn't in task. So, I will try it...
p_(H2O)V = n_(H2O)RT
p(H2O) = n_(H2O)RT/V = 2,52 * 8,31 * (100+273,15)/0,05 Pa = 156 284 Pa

 

[Borek]

It must be a partial pressure, as not all gases reacted (hydrogen was in excess).

If there is a condensed water present, what MUST be water vapor pressure at 100 deg C?

Or, from slightly different angle - what typically happens to water at 100 deg C?

 

[Chestermiller]

The partial pressure of water vapor = pressure of water vapor in the V of the system and with temperature T system... Is it right? However, why particial pressure? It isn't in task. So, I will try it...
p_(H2O)V = n_(H2O)RT
p(H2O) = n_(H2O)RT/V = 2,52 * 8,31 * (100+273,15)/0,05 Pa = 156 284 Pa

No. This is not correct for the reasons that Borek stated. In addition, the gas does not fill the entire container, so that the volume of the gas is not 0.05 m^3. There is some liquid water in the container that takes up part of the volume now. So you can't substitute 0.05 into the ideal gas law once the liquid water is present. We will get to all that in due course. So again, to reiterate Borek's question, what is the vapor pressure of water at 100C?

Chet

 

[Borek]

You can safely ignore the liquid volume, it is orders of magnitude lower than the volume of the gas.

 

[Chestermiller]

You can safely ignore the liquid volume, it is orders of magnitude lower than the volume of the gas.

This is true for this particular problem. But, the OP doesn't know that yet. I think it would be helpful for him, in terms of his learning experience, to not automatically assume this yet. The equations will certainly reveal this.

Chet

 

[Numeriprimi]

Water vaporizes typically at 100° C of its entire volume - atmospheric pressure = vapor pressure = 101 215 Pa... This answer?

 

[Chestermiller]

Water vaporizes typically at 100° C of its entire volume - atmospheric pressure = vapor pressure = 101 215 Pa... This answer?

Yes. This is correct. So, let's review where we stand. We have a 50 liter container at 100 C. The container contains 2 chemical species, H2 and H2O (2.38 moles, and 2.52 moles). The water is present in 2 phases, a liquid phase and a gaseous phase. All the H2 is in the gaseous phase. The partial pressure of the H2O in the gas phase is 101215 Pa. The partial pressure of the H2 in the gas phase is currently unknown. The molar split between liquid water and water vapor is also unknown. These are what we want to find.

Let x = number of moles of water in the gas phase
Let (2.52-x)= number of moles of liquid water

In terms of x, how many grams of liquid water are there in the container?
In terms of x, what is the volume of liquid water in the container?
In terms of x, what is the volume of gas in the container?
Knowing the partial pressure of water vapor in the container, the volume of gas in the container (in terms of x), the number of moles of water in the gas phase (in terms of x), and the temperature, use the ideal gas law to determine x. How many grams of liquid water are in the container. How many cc's? What fraction of the total 50 liter volume is this?

After you complete this part, we can next focus on the partial pressure of the H2.

Chet

 

[Numeriprimi]

In terms of x, how many grams of liquid water are there in the container?
n_l = 2.52-x = m_l/M_H2O; m_l = (2.52-x)*(2*0,00202+0,032) kg = 36,04 * (2.52-x) g
In terms of x, what is the volume of liquid water in the container?
V_l = m_l/ρ = 0.036 04*(2.52-x)/1000 m^3
In terms of x, what is the volume of gas in the container?
V_g = V - V_l = (0.05 - 0.036 04*(2.52-x)/1000) m^3

pV=nRT
101 215*(0.05-(0.036 04*(2.52-x)/1000)) = (2.52-x)*8.31*373.15
x = 0.9 mol is the number of water in the gas phase; (2,52 - 0,9)mol = 1,62 mol
m = M_H20*x = (2*0.00202+0.032)*1,62 = 0. 058 kg 7 (V= m/M_M * V_M = 0,058/0,03604*0,022414 m3 = 0,036m3 - 72 %)

 

[Chestermiller]

In terms of x, how many grams of liquid water are there in the container?
n_l = 2.52-x = m_l/M_H2O; m_l = (2.52-x)*(2*0,00202+0,032) kg = 36,04 * (2.52-x) g

The molecular weight of water is 18, not 36. Otherwise, what you've done is correct. Nice job. Please correct this and subsequent relationships.

In terms of x, what is the volume of liquid water in the container?
V_l = m_l/ρ = 0.036 04*(2.52-x)/1000 m^3
In terms of x, what is the volume of gas in the container?
V_g = V - V_l = (0.05 - 0.036 04*(2.52-x)/1000) m^3

The above part will be correct once you correct the molecular weight mistake.

pV=nRT
101 215*(0.05-(0.036 04*(2.52-x)/1000)) = (2.52-x)*8.31*373.15

There is an error in the above equation. You need to use the number of moles of water in the gas phase x on the right hand side, rather than the number of moles of water in the liquid phase. The ideal gas law is for the gas phase. Please correct this and try again.

Chet

 

[Numeriprimi]

Ok, yees, because the M_MO = 32 g/mol for O_2, not O, so I have to divide twice and add 2 g/mol for H_2...

In terms of x, how many grams of liquid water are there in the container?
n_l = 2.52-x = m_l/M_H2O; m_l = 18 * (2.52-x) g
In terms of x, what is the volume of liquid water in the container?
V_l = m_l/ρ = 0.018*(2.52-x)/1000 m^3
In terms of x, what is the volume of gas in the container?
V_g = V - V_l = (0.05 - 0.018*(2.52-x)/1000) m^3

pV=nRT
101 215*(0.05 - 0.018*(2.52-x)/1000) = x*8.31*373.15
So, x = 1,63152 mol is the number of moles of water in the gas phase
m = M_H20*x = 0, 018*1,63152 kg = 0, 029 37 kg
V= n*V_m = 1,63152*0,022414 m3 = 0, 03657 m3 (73 %)
It is OK now?

 

[Chestermiller]

Ok, yees, because the M_MO = 32 g/mol for O_2, not O, so I have to divide twice and add 2 g/mol for H_2...

In terms of x, how many grams of liquid water are there in the container?
n_l = 2.52-x = m_l/M_H2O; m_l = 18 * (2.52-x) g
In terms of x, what is the volume of liquid water in the container?
V_l = m_l/ρ = 0.018*(2.52-x)/1000 m^3
In terms of x, what is the volume of gas in the container?
V_g = V - V_l = (0.05 - 0.018*(2.52-x)/1000) m^3

pV=nRT
101 215*(0.05 - 0.018*(2.52-x)/1000) = x*8.31*373.15
So, x = 1,63152 mol is the number of moles of water in the gas phase

The above looks OK now.

m = M_H20*x = 0, 018*1,63152 kg = 0, 029 37 kg
V= n*V_m = 1,63152*0,022414 m3 = 0, 03657 m3 (73 %)
It is OK now?

I can't figure out what you did here. The m you calculated of 29 g is the mass of water in the gas phase. The number of moles of liquid water is (2.52-1.63)=0.89. The mass of the liquid water is 0.89x18=16 grams. The volume occupied by the liquid water is 16 cc. This is 0.032% of the total 50 liter volume. So the volume of the gas is 49984 cc = 49.984 liters. This confirms Borek's comment regarding the volume of the liquid water in the container compared to the total volume of the container.

Based on this volume, the number of moles of H2 in the gas, and the temperature, you can use the ideal gas law to calculate the partial pressure of H2 in the gas phase. What partial pressure do you get? What is the sum of the partial pressures of H2 and H2O (which is the total pressure in the container)?
Chet

 

[Numeriprimi]

Ok, thank you :-)

pV=nRT
p_H = nRT/V = 4,9*8,31*373,15/0,049984 Pa = 303 983 Pa
The sum of partial pressures of H2 and H2O is 303 983 Pa + 101 215 Pa = 405 198 Pa

 

[Chestermiller]

Ok, thank you :-)

pV=nRT
p_H = nRT/V = 4,9*8,31*373,15/0,049984 Pa = 303 983 Pa
The sum of partial pressures of H2 and H2O is 303 983 Pa + 101 215 Pa = 405 198 Pa

Very nice. I like your way of getting the total pressure better, except that the total number of moles of gas is only 4.01 = (2.38 + 1.63)

Chet