# Question #d1755

##### 1 Answer

#### Answer:

#### Explanation:

The trick here is to realize that the **only** thing that changes for gas **volume** it occupies.

The amount of gas *number of moles* of gas

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 **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

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

#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.