A person evacuated a cylinder and filled in $4~\mathrm{g}$ of gas A at $25~\mathrm{^\circ C}$. Pressure was found to be $1~\mathrm{atm}$. Another person filled another $8~\mathrm{g}$ of gas B in the cylinder at the same temperature. The final pressure was found to be $1.5~\mathrm{atm}$. What is the ratio of molecular masses of A and B (assuming ideal gas behaviour)?

I tried to do it this way:

Taking the molecular mass of A as $x\ \mathrm{g}$ and that of B as $y\ \mathrm{g}$, I formed the respective ideal gas equations but the required terms ($x$ and $y$) just cancelled out.

$$N(a) = 4/x ;\quad N (b) = 8/y$$
So$$V (a) = \frac{4 \times 22.4}{x};\quad V (b) = \frac{8 \times 22.4}{y}$$

So, ideal gas equation for A: $$1 X \frac{4 \times 22.4}{x} = \frac {4 \times R \times 25} x$$ Here $x$ cancels out. Similarly $y$ cancels out in its equation.
This is the problem I am facing.

Another thing I want to know is that the total pressure (i.e $1.5~\mathrm{atm}$), will that be the sum of the individual pressures of gases A and B?

  • 1
    $\begingroup$ Could you edit your question to show the math whereby your terms cancel out? $\endgroup$
    – airhuff
    Feb 10 '17 at 15:54
  • $\begingroup$ To answer your last question, yes, that is a central concept of the ideal gas law; that the composition of the gases is irrelevant with respect to the total pressure of the system. Regarding the rest of your question, don't forget units! That will both further clarify what you are doing and help you make sense of it yourself. If the units don't make sense, your equation probably doesn't, and vise versa. $\endgroup$
    – airhuff
    Feb 10 '17 at 16:31
  • $\begingroup$ @airhuff units will only determine the quantity involved. But to solve we need the numerics. $\endgroup$
    – Saksham
    Feb 10 '17 at 16:52
  • $\begingroup$ I don’t quite get where and how parts of this interact with the ideal gas equation, and I don’t understand the last equation block (is there something missing where the $1$ is?) but at least you showed your work so it’s no longer homework. $\endgroup$
    – Jan
    Feb 10 '17 at 16:53
  • $\begingroup$ @Jan Yes, it's not a part of my homework. The 1 is the pressure found when gas A is put in the cylinder. So that 1 is just being multiplied. It was a question based on the ideal gas equation so it has to be solved that way (according to me). But if u have any other simple method, please let me know. $\endgroup$
    – Saksham
    Feb 10 '17 at 16:55

The ideal gas equation is $PV = nRT$

But for this problem V, R, and T are constants. So we can collect terms as:

$\dfrac{P}{n} = \dfrac{RT}{V}$

So $\dfrac{P_A}{n_A} = \dfrac{P_B}{n_B}$

Now you correctly determined the number of moles of A and B.

$n_A = \dfrac{4}{\text{MW}_A}$
$n_B = \dfrac{8}{\text{MW}_B}$

But you need to consider the partial pressures of A and B.

The cylinder was empty before A was added, and after A was added the pressure was 1 atmosphere. So the partial pressure of A is 1 atmosphere.

Before B is added the pressure is 1 atmosphere and after B is added the pressure is 1.5 atmospheres. So the partial pressure of B is 0.5 atmospheres.

$P_A = 1$
$P_B = 0.5$

Now substituting into our simplified equation

$\dfrac{1}{\frac{4}{\text{MW}_A}} = \dfrac{0.5}{\frac{8}{\text{MW}_B}}$

$\dfrac{8}{\text{MW}_B} = \dfrac{2}{\text{MW}_A}$

$\dfrac{\text{MW}_A}{\text{MW}_B} = \dfrac{2}{8} = \dfrac{1}{4}$

  • $\begingroup$ Everything's fine but I still can't understand why the volume is a constant. $\endgroup$
    – Saksham
    Feb 10 '17 at 17:39
  • 1
    $\begingroup$ Same cylinder. We're assuming that the pressure doesn't deform the cylinder. Obviously if the "cylinder" was a balloon then the volume would change as the pressure changes. $\endgroup$
    – MaxW
    Feb 10 '17 at 17:42

Ideal gas equation is PV = nRT Here R is constant temperature is constant mentioned in question and also the volume is constant as volume of gas equals to the volume of cylinder (volume of gas= space occupied by it)

So in ideal gas equation n and P are variables rest are constant equation can be modified as,

P1/n1 = P2/n2

P1=1 n1=4/x n2 P2=1.5

By solving we get n2=6/x

It is total moles of gas filled in cylinder i.e. 4/x + 8/y

6/x = 4/x +8/y Multiplying equation by x we get 6=4+8x/y

x/y = 1/4

  • $\begingroup$ When you say that the volume of gas equals the volume of the cylinder, do you mean the cumulative or combined volume of both the gases A and B? $\endgroup$
    – Saksham
    Feb 10 '17 at 17:35
  • $\begingroup$ Moreover, the P2 which you have written as equal to 1.5atm, that is the combined pressure of both the gases when both of them are present in the cylinder simultaneously, and not only of gas B as you seem to suggest. $\endgroup$
    – Saksham
    Feb 10 '17 at 17:37
  • $\begingroup$ Cumulative as in Q. Cylinder dimensions aren't mentioned so for sake of question we must assume that both gases are filled in cylinder of same dimension and hence have same volume $\endgroup$
    – user41111
    Feb 10 '17 at 17:40
  • $\begingroup$ OK ok it's my mistake everything is ok but p2 will have different value $\endgroup$
    – user41111
    Feb 10 '17 at 17:44
  • $\begingroup$ Ya that's why I was getting confused. Thanks anyways. $\endgroup$
    – Saksham
    Feb 10 '17 at 17:47

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