Answer:
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.
Step by Step Explanation?
Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.
The formula is,
P₁ x V₁ = P₂ × V₂
Where,
P₁ is initial pressure = 1 atm
P2 is final pressure = ? (Not Known)
V₁ is initial volume = 10 L
V₂ is final volume = 15 L
Now put the values in the formula,
\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]
Therefore, the answer is 0.67 atm.
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)

x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Unless you are talking about one specific theory, the answer is pseudoscience.
No, because hydrogen isn’t brought out of the equation