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

Answer:

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

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 .

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

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

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.