Question

A gas sample containing a constant number of gas molecules has a volume of 2.70 L at a constant pressure and a temperature of 25.0o C. What would be the volume (in Liters) of this gas sample at 75.0o C? Round your answer to 3 sig fig

Answer 1

`\boxed {\boxed {\sf 8.10 \ L}}`

Step-by-step explanation:

This question asks us find the volume of a gas sample given a change in temperature. Since the pressure remains constant, we only are concerned with the variables of temperature and volume.

We will use Charles's Law. This states the volume of a gas is directly proportional to the temperature of a gas. The formula is:

`(V_1)/(T_1)=(V_2)/(T_2)`

The gas starts at a volume of 2.70 liters and a temperature of 25.0 degrees Celsius.

`\frac {2.70 \ L}{25.0 \textdegree C}=(V_2)/(T_2)`

The temperature is increased to 75.0 degrees Celsius, but the volume is unknown.

`\frac {2.70 \ L}{25.0 \textdegree C}=(V_2)/(75.0 \textdegree C)`

We are solving for the volume at 75 degrees Celsius, so we must isolate the variable V₂.

It is being divided by 75.0 °C. The inverse operation of division is multiplication, so we multiply both sides of the equation by 75.0 °C.

`75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}=(V_2)/(75.0 \textdegree C) * 75.0 \textdegree C`

`75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}= V_2`

The units of degrees Celsius (° C) cancel.

`75.0 *\frac {2.70 \ L}{25.0}= V_2`

`75.0 *0.108 \ L = V_2`

`8.1 \ L = V_2`

The original measurements have 3 significant figures, so our answer must have the same. Currently, the answer has 2. If we add another 0, the value of the answer does not change, but the number of sig figs does.

`8.10 \ L = V_2`

The volume of this gas sample at 75.0 degrees Celsius is 8.10 Liters.