Question

A balloon contains 20.0 L of helium in the morning when the temperature is

23.0°C. By afternoon, the temperature has increases to 45.0°C. What is the new
volume of the balloon? *
O 21.49 L
39.13 L
0 0.02 L

Answer 1

`\boxed {\boxed {\sf 39.13 \ Liters}}`

Step-by-step explanation:

We want to find the new volume given the temperature, so we use Charles's Law. This states the volume of a gas is directly proportional to the temperature. The formula is:

`\frac {V_1}{T_1}=(V_2)/(T_2)`

We know the original volume is 20.0 liters and the temperature is 23.0 degrees Celsius. The temperature changes to 45.0 degrees Celsius, but we don't know the new volume. Substitute the known values into the formula.

`\frac {20.0 \ L}{23.0 \textdegree C}=( V_2)/(45.0 \textdegree C)`

We are solving the new volume, so we need to isolate the variable V₂. It is being divided by 45.0 degrees Celsius. The inverse of division is multiplication, so we multiply both sides by 45.0 °C.

`45 \textdegree C *\frac {20.0 \ L}{23.0 \textdegree C}=( V_2)/(45.0 \textdegree C)* 45 \textdegree C`

`45 \textdegree C *\frac {20.0 \ L}{23.0 \textdegree C} = V_2`

The units of degrees Celsius cancel.

`45 *\frac {20.0 \ L}{23.0}= V_2`

`39.1304348 \ L=V_2`

If we round to the nearest hundredth, the 0 in the thousandth place tells us to leave the 3.

`39.13 \ L \approx V_2`

The new volume of the balloon is approximately 39.13 Liters.