If #3 L# of a gas at room temperature exerts a pressure of #25 kPa# on its container, what pressure will the gas exert if the container's volume changes to #5 L#?

2 Answers
Jul 3, 2017

Answer:

The new pressure is #=15kPa#

Explanation:

We apply Boyle's Law

#P_1V_1=P_2V_2#

The initial pressure is #P_1=25kPa#

The initial volume is #V_1=3L#

The final volume is #V_2=5L#

The final pressure is

#P_2=V_1/V_2*P_1#

#=3/5*25=15kPa#

Jul 3, 2017

Answer:

The final volume will be #"15 kPa"#.

Explanation:

This question involves Boyle's law, which states that the volume of a given amount of gas held at constant temperature, varies inversely with the applied pressure. This means that if the pressure increases, the volume decreases, and vice-versa. The equation for this law is:

#P_1V_1=P_2V_2#

Known

#P_1="25 kPa"#

#V_1="3 L"#

#V_2="5 L"#

Unknown

#P_2#

Solution

Rearrange the equation to isolate #P_2#. Substitute your data into the equation and solve.

#P_2=(P_1V_1)/(V_2)#

#P_2=(25"kPa"xxcolor(red)cancel(color(black)("3L")))/(5color(red)cancel(color(black)("L")))="15 kPa"#