Question

A sample of gas in a balloon has an initial temperature of 23 ∘C and a volume of 1.09×103 L . If the temperature changes to 59 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

Answer 1

`1.22\cdot 10^3 L`

Step-by-step explanation:

We can solve the problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas is directly proportional to its temperature:

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

where here we have

`V_1 = 1.09\cdot 10^3 L` is the initial volume

`T_1 = 23^(\circ)+ 273 = 296 K` is the initial temperature

`V_2` is the final volume

`T_2 = 59^(\circ)+ 273 =332 K` is the final temperature

Solving for V2, we find

`V_2 = (V_1 T_2)/(T_1)=((1.09 \cdot 10^3 L)(332 K))/(296 K)=1.22\cdot 10^3 L`