What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 42 ∘C , V1= 1.10×103 L ) to a volume of 2.20×103 L ? Assume no change in pressure or the amount of gas in the balloon.

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is maintained at a constant pressure, by means of a constant of proportionality that is applied directly. For a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement of the gas molecules. .

In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between the volume and the temperature will always have the same value:

[tex]\frac{V}{T} =k[/tex]

Now it is possible to assume that you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

[tex]\frac{V1}{T1} =\frac{V2}{T2}[/tex]

In this case:

V1= 1.10*10³ L
T1= 42 C= 315 K (being 0 C=273 K)
V2= 2.20*10³ L
T2= ?

Replacing:

[tex]\frac{1.10*10^{3}L }{315 K} =\frac{2.20*10^{3}L }{T2}[/tex]

Solving:

[tex]T2=\frac{2.20*10^{3}L }{ \frac{1.10*10^{3}L }{315 K} }[/tex]

T2=630 K= 357 C

The temperature that is required is 357 C.