Question

4.
Ammonia gas occupies a volume of 450. mL at a pressure of 720 mm Hg. What volume in
liters will the gas occupy at standard atmospheric pressure?

Answer 1

P1V1 = P2V2

P1 = 720 mmHg
V1 = 450. mL
P2 = 760 mmHg (this is the pressure at STP)

Use these to solve for V2:
(720)(450) = 760V2

V2 = 426 mL

Answer 2

Answer:

$\boxed {\boxed {\sf 426 \ mL}}$

Explanation:

We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:

$P_1V_1= P_2V_2$

The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.

$450 \ mL * 720 \ mm \ Hg = P_2V_2$

The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.

$450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2$

We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.

$\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}$

$\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2$

The units of millimeters of mercury (mm Hg) cancel.

$\frac {450 \ mL * 720 }{760} = V_2$

$\frac {324,000}{760} \ mL = V_2$

$426.3157895 \ mL =V_2$

The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.

$426 \ mL \approx V_2$

The volume at standard atmospheric pressure is approximately 426 milliliters.