Question

A student combined 1.2 g of H2 gas and 5.4 g of He gas in a 2.5 L closed container. What was the pressure exerted by the H2 gas if the mixture was at 22°C? (R = 0.08206 L · atm/mol · K)

Answer 1

You can do it this way...
You first take the Mole of these gases

First lets figure out the Hydrogen gas so for that it will be:
\[Mass \div Mr= 1.2 \div 2 = 0.6\]

Answer 2

So now you use the formula for
\[Pv=nRT \]

Where P is pressure
v is Volume
n is Mole
R is the constant value
T is temperature

there are certain things in this question that needs to be converted to standard unit.
Which in our question is T,
So for T, which is 22°C is suppose to be changed from °C to K (Kelvin)
how do convert it? simple! just add 273 to it. In our case it will be 295 (273 + 22).

Answer 3

For the R value thats already been given to you is
R = 0.0821 L atm K-1 mol-1 which if you break it down it will be like this:

Pressure is in atmospheres(atm), Volume is in litres(L), Temperature is in Kelvin(K), which is basically the not the value we used for standard unit

what you are suppose to consider the R value as is 8.314 J K-1 mol-1

because the unit for this R value is Pressure is in atmospheres(atm), Volume is in litres(L), Temperature is in Kelvin(K) which IS considered to be standard unit!

Answer 4

So now you have the R value, T value, n Value only one more thing you need to figure out is the v Value. So what you are given is L (which is 1000cm3 ) in order to make it in to a into startdard value you need to make it into m3

cm3 to m3 is by dividing it by 1 million
which will be 2.5 x 10^-6

Answer 5

Now you have all the values! now just plug in the values! :)

T= 295K
R = 8.31
n = 0.6
v = 2.5 x 10^-6

\[Pv=nRT\]

\[P= 0.6 \times 8.31 \times 295 \]
\[P = 1470 \div (2.5 \times 10^{-6})\]
\[P= 5.8 \times 10^{8}\]

Answer 6

So now the value for P should be Pa so it's:

5.8 x 10^8 Pa
Btw the reason why you do not take the Mole value of He because they are not even asking for it!!! it's just to confuse you so you just ignore He.

I hope this cleared this it out for you. if you have anymore question, don't hesitate to ask.

:)

Answer 7

if you need the value to be atm you can easily convert it from Pa to atm too :)