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If #19 L# of a gas at room temperature exerts a pressure of #2 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12 L#?

Answer 1

Isabella Alvarez

#P_2=3,17 kPa#

#P_1=2kPa# #V_1=19L# #V_2=12L# #P_2=?#

#"formula :"#

#P_1*V_1=P_2*V_2" Boyle's law"#

#2*19=P_2*12#

#P_2=(2*19)/12#

#P_2=3,17 kPa#

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Answer 2

Axel Acevedo

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

Boyle's Law formula:

P1 * V1 = P2 * V2

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

Given: P1 = 2 kPa V1 = 19 L V2 = 12 L

We can plug in the values and solve for P2:

P2 = (P1 * V1) / V2

P2 = (2 kPa * 19 L) / 12 L

P2 = 38 kPa

Therefore, the gas will exert a pressure of 38 kPa when the container's volume changes to 12 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.