Question #d1755

The trick here is to realize that the only thing that changes for gas #"B"# after the valve is opened is the volume it occupies.

The amount of gas #"B"# present in the vessel does not change, i.e. the number of moles of gas #"B"# remains constant.

As you know, when temperature and number of moles of gas are kept constant, the pressure exerted by a gas is inversely proportional to the volume it occupies #-># this is known as Boyle's Law.

Under these conditions, increasing the volume of the gas will cause its pressure to decrease. Similarly, decreasing the volume will cause the pressure to increase.

Mathematically, this can be written as

#color(blue)(ul(color(black)(P_1 * V_1 = P_2 * V_2)))#

Here

• #P_1# and #V_1# are the pressure and volume of the gas at an initial state
• #P_2# and #V_2# are the pressure and volume of the gas at a final state

In your case, you know that

#{(P_1 = "630 torr"), (V_1 = "82 L") :}#

When you open the valve, the volume of the gas will increase to include the volume of the bulb in which gas #"A"# was being kept.

You can thus say that after the valve is opened, the volume of the flask will be

#V_"flask" = "59 L" + "82 L" = "141 L"#

Therefore, the partial pressure of gas #"B"# after the valve is opened

#P_1 * V_1 = P_2 * V_2 implies P_2 = V_1/V_2 * P_1#

will be equal to

#P_2 = (82 color(red)(cancel(color(black)("L"))))/(141color(red)(cancel(color(black)("L")))) * "630 torr" = color(darkgreen)(ul(color(black)("370 torr")))#

The answer is rounded to two sig figs.

As you can see, increasing the volume of the gas caused the pressure to decrease, just as we would expect.