a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a factor of 2 what will happen to the pressure? you may assume the temperature remains at 35 c

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

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