Home > Physics > Gas Laws

If #14/3 L# of a gas at room temperature exerts a pressure of #45 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12/5 L#?

Answer 1

Lily Adams

The pressure is #=87.5kPa#

Boyle's Law is employed.

#P_1V_1=P_2V_2#

#P_1=45kPa#

#V_1=14/3 L#

#V_2=12/5L#

#P_2=(P_1V_1)/V_2=(45*14/3)/(12/5)=87.5kPa#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Amelia Adams

Using Boyle's Law, we can solve for the new pressure:

P1 * V1 = P2 * V2

Where: P1 = initial pressure = 45 kPa V1 = initial volume = 14/3 L P2 = final pressure (what we're solving for) V2 = final volume = 12/5 L

Substituting the given values:

45 kPa * (14/3 L) = P2 * (12/5 L)

Solving for P2:

P2 = (45 kPa * (14/3 L)) / (12/5 L) P2 = (45 * 14) / (3 * 12/5) P2 = (630) / (36/5) P2 = 875 kPa

So, the gas will exert a pressure of 875 kPa when the container's volume changes to 12/5 L.

By signing up, you agree to our Terms of Service and Privacy Policy

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.