Respuesta :
Answer:
If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.
Explanation:
Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure
V ∝ 1/P or V₁·P₁ = V₂·P₂
Where:
V₁ = Initial volume
V₂ = Final volume = V₁/2
P₁ = Initial pressure = 832 torr
P₂ = Final pressure = Required
From V₁·P₁ = V₂·P₂ we have,
P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)
P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr
Answer:
The pressure will increase by a factor of 2 and is now 1664 torr
Explanation:
The question says for us to assume that temperature is constant. Now, since we are given pressure and volumw, we will use Boyle's law which is a law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature i.e V ∝ 1/P
Thus, PV = K
Where K is a constant.
So,
P₁•V₁ = P₂•V₂ = k
Where:
P₁ = Initial pressure
V₁ = Initial volume
P₂ = Final pressure
V₂ = Final volume = V₁/2
From the question, P₁ = 832 torr ; we are told that volume is decreased by 2,thus, V₂ = V₁/2
Now, we want to find out what will happen to the pressure P₂.
Let's make P₂ the subject;
P₁•V₁ = P₂•V₂
Thus, P₂ = (P₁•V₁)/V₂
Plugging in the relevant values to obtain ;
P₂ = (P₁•2V₁)/V₁
V₁ will cancel out and we have;
P₂ = 2P₁
Now, we are given that P₁ = 832 torr, Thus,
P₂ = 2P₁ = 2× 832 torr = 1664 torr
