Old Dalton's law of partial pressures states that in a gaseous mixture, the exerted by a component gas is the same as the pressure it would exert IF IT ALONE occupied the container.....
And thus....$P_"Total"=P_(N_2)+P_(O_2)+P_(H_2)$....
$80%$of pressure is exerted by nitrogen gas. So, $20%$of pressure will be exerted by oxygen gas. $"p"_("O"_2) = 20% "of 2 atm" = 20/100 × "2 atm" = "0.4 atm"$
And so $P_(O_2)=P_"Total"-P_(He)-P_(CO_2)$ $={101.4-82.5-0.4}*kPa$ $=18.5*kPa$ In commercial diving mixes I am not sure that carbon dioxide would be added in those quantities. I used to do a lot of recreational...
The total pressure is the sum of the partial pressures of the given gases. $P_"Total"=P_"He"+P_"CO2"+P_"O2"$ Rearrange the equation to isolate $P_"O2"$, substitute the given values into the equation and solve....
Datlon's law of Partial Pressures states that in a gaseous mixture, the exerted by a gaseous component is the same as the pressure it would exert if it ALONE occupied...
Dalton's law of partial pressures states that in a gaseous mixture, the of a gaseous component is the same as the pressure that gas would exert if it ALONE occupied...
$"Dalton's Law of partial pressures"$ states that in a gaseous mixture, the exerted by any component is the same the pressure it would exert if it alone occupied the container....
According to Dalton's law of partial pressures, the total pressure is equivalent to the sum of the pressures of the individual gases (if they were present alone). Therefore, the...
And so... $P_"Total"=P_"He"+P_"Ne"+P_"Ar"=490*mm*Hg$ But we are given that $P_"He"=21.5*mm*Hg$, and that $P_"Ar"=10.2*mm*Hg$.. And so...$P_"Ne"={490-21.5-10.2}*mm*Hg}=458.3*mm*Hg$..
This shouldn't feel too complicated... We know: what the vapor pressure of water vapor plus hydrogen gas is. what the vapor pressure of just water vapor is. Hence,...
From Dalton's Law of , the total pressure of a mixture of gases is the sum of the partial pressures of each gas. Therefore, the partial pressure of $"CO"_2$ is...