Question

A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L , what is the new pressure?

Answer 1

`\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}`

Step-by-step explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

`P_1V_1=P_2V_2`

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

`25.0 \ L * 2.05 \ atm = P_2V_2`

The volume is decreased to 14.5 liters, but the pressure is unknown.

`25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L`

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

`\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=(P_2 *14.5 \ L)/(14.5 \ L)`

`\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2`

The units of liters cancel.

`\frac {25.0 * 2.05 \ atm }{14.5 }=P_2`

`\frac {50.25\ atm }{14.5 }=P_2`

`3.53448276 \ atm = P_2`

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

• 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

`3.53 \ atm \approx P_2`

The new pressure is approximately 3.53 atmospheres.