Question

container x contains 1.0 mol of an ideal gas. container y contains 2.0 mol of the ideal gas. y has four times the volume of x. the pressure in x is twice that in y. What is Temperature of gas in X/temperature of gas in Y.

Answer 1

Answer:

Correct option is C)

Ideal gas equation PV=nRT.

For gas X: P=P,V=2V,(n=m/a);T=200K.

Final position: P=P,V=V,(n=m

2

/a),T=400K.

By comparing them, m

2

=(m/4).

Answer 2

Answer:

$\frac{T_x}{T_y}$ = 1

Explanation:

This is your best friend formula:

$PV=nRT$

Now we need two of these, for Container X and Container Y.

$P_x V_x = n_xRT_x\\P_y V_y = n_yRT_y$

Notice how R is the same on both of them? R is the Ideal Gas Constant (which is 8.31Jmol^-1K^-1)... which is the same for both containers.

Rearrange the equation for R on both and you notice that you can do R=R

$\frac{P_x V_x}{n_xT_x} = \frac{P_y V_y}{n_yT_y}$ twins eh?

Now you can reread the question. $n_x$ basically disappears because it = 1 in the denominator. Almost the same for $n_y$ but it turns into a 2.

$\frac{P_x V_x}{T_x} = \frac{P_y V_y}{2T_y}$

Since the question is asking for the ratio  $\frac{T_x}{T_y}$, you can rearrange the equation to get that.

$\frac{T_x}{T_y} = \frac{2P_xV_x}{P_yV_y}$ This is something you should get.

Y has four times the volume of X. So Y is the big boi one, and you need 4 of the X container volumes to make up 1 Y Container volume. Not the other way around, some can get tricked by this! $4V_x = V_y$

Pressure in X is twice that in Y. Sentence structure is weird on this one. Say it as "Pressure in X is double that in Y". So X is the big boi one here, and you need Double Y to make up 1 X. So $P_x = 2P_y$.

Now we got $V_y$ and $P_x$ to use as weapons, substitute that into the rearranged question.

$\frac{T_x}{T_y} = \frac{2(2P_y)V_x}{P_y(4V_x)} = \frac{4P_yV_x}{P_y4V_x}$

Aha! $P_y$ gets cancelled. $V_x$ gets destroyed. $4$ gets executed!

And all you have is a delicious 1.