Question

A balloon containing helium gas has a volume of 1.25 L at room temperature (23 oC). The balloon is heated to at temperature of 85 oC . Assuming no change in pressure, what is the new volume of the balloon?

Answer 1

Answer:

$\boxed {\boxed {\sf 4.6 \ Liters}}$

Explanation:

The pressure stays constant, so we are dealing with volume and temperature, so we use Charles's Law. This states the temperature and volume of a gas are directly proportional. The formula is:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

We know the original balloon has a volume of 1.25 liters at a temperature of 23 degrees celsius. These values can be substituted in.

$\frac{1.25 \ L}{23 \textdegree C}=\frac{V_2}{T_2}$

The new volume is unknown, but the temperature is increased to 85 degrees Celsius.

$\frac{1.25 \ L}{23 \textdegree C}=\frac{V_2}{85 \textdegree C}$

We are trying to solve for the new volume, V₂. It is being divided by 85 degrees Celsius. The inverse of division is multiplication, so we multiply both sides by 85°C.

$85 \textdegree C*\frac{1.25 \ L}{23 \textdegree C}=\frac{V_2}{85 \textdegree C}*85 \textdegree C$

$85 \textdegree C*\frac{1.25 \ L}{23 \textdegree C}= V_2$

The units of degrees Celsius cancel.

$85 *\frac{1.25 \ L}{23} = V_2$

$4.61956522 \ L = V_2$

The original measurements have at least 2 significant figures, so our answer must have 2. For the number we found, that is the tenth place.

• 4.61956522

The 1 in the hundredth place (in bold above) tells us to leave the 6 in the tenth place.

$4.6 \ L \approx V_2$

The new volume of the balloon is approximately 4.6 liters.