# Question #c925d

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that the volume of a gas is **Inversely proportional** to its pressure when the number of moles of gas and the temperature of the gas are kept constant **Boyle's Law** here.

This basically means that when the volume of a gas **decreases**, its pressure **increases** by the same factor.

#"volume" darr implies "pressure" uarr#

Mathematically, this is expressed using the following equation

#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#

Here

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

In your case, the volume is **decreasing**

#"50 cm"^3 " " darr " " "20 cm"^3#

so you should expect the pressure of the gas, which would be *increase* as a result

#"0.49346 atm" " "uarr " " P_2#

Rearrange the equation to solve for

#P_1V_1 = P_2V_2 implies P_2 = V_1/V_2 * P_1#

Plug in your values to find

#P_2 = (50color(red)(cancel(color(black)("cm"^3))))/(20color(red)(cancel(color(black)("cm"^3)))) * "0.49346 atm"#

#P_2 = "1.23365 atm"#

Finally, to convert this to *kPa*, use the fact that

#"1 atm = 101.325 kPa"#

You will end up with

#1.23365 color(red)(cancel(color(black)("atm"))) * "101.325 kPa"/(1color(red)(cancel(color(black)("atm")))) = "125 kPa" = color(darkgreen)(ul(color(black)("100 kPa")))#

The answer **must** be rounded to one **significant figure**, the number of sig figs you have for the two volumes.