Home > Physics > Gas Laws

If #19 L# of a gas at room temperature exerts a pressure of #4 kPa# on its container, what pressure will the gas exert if the container's volume changes to #7 L#?

Answer 1

Benjamin Asiedu

#11*10^3 Pa#

When the containers volume changes to #7 L# from #19 L# some work must be done hence from Boyle's law We can say that

#P_1V_1=P_2V_2#

where #PV="costant"#

thus by changing the values, we obtain

#19*4*10^3=7*P_2#

#P_2=10.9*10^3~=11*10^3#

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

Answer 2

Parker Abbott

To find the new pressure, you can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume at constant temperature. Using the formula P1V1 = P2V2, where P1 is the initial pressure (4 kPa), V1 is the initial volume (19 L), and V2 is the final volume (7 L), you can solve for P2:

P2 = (P1 * V1) / V2 P2 = (4 kPa * 19 L) / 7 L P2 ≈ 10.86 kPa

Therefore, the gas will exert a pressure of approximately 10.86 kPa when the container's volume changes to 7 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.