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If #8 L# of a gas at room temperature exerts a pressure of #32 kPa# on its container, what pressure will the gas exert if it the container's volume changes to #4 L#?

Answer 1

Elena Albarran

#32kPa#

Since the temperature is constant in this instance, the ideal gas law states:

#P_1V_1=P_2V_2#

#or,8Lxx32kPa=4LxxP_2#

#or,P_2=(8Lxx32kPa)/(4L)#

#:.P_2=64kPacolor(brown)#

Therefore, we must apply pressure of (64-32) kPa or 32 kPa.

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Answer 2

Benjamin Asiedu

To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume, provided the temperature remains constant.

The formula for Boyle's Law is:

P₁V₁ = P₂V₂

Where: P₁ = initial pressure V₁ = initial volume P₂ = final pressure V₂ = final volume

Given: P₁ = 32 kPa V₁ = 8 L V₂ = 4 L

Substitute the values into the formula:

P₁V₁ = P₂V₂

32 kPa * 8 L = P₂ * 4 L

Solve for P₂:

P₂ = (32 kPa * 8 L) / 4 L P₂ = 256 kPa

Therefore, the gas will exert a pressure of 256 kPa when the container's volume changes to 4 L.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.