Question

A sample of gas in a balloon has an initial temperature of 20. ∘C and a volume of 1.92×103 L . If the temperature changes to 68 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

Answer 1

`\boxed{2.23 * 10^(3) \text{ L}}`

Step-by-step explanation:

The pressure is constant, so we can use Charles' Law.

`( V_(1) )/(T_(1)) = ( V_(2) )/(T_(2))`

Data:

V₁ = 1.92 × 10³ L; T₁ = 20 °C

V₂ = ?; T₂ = 68 °C

Calculations:

(a) Convert temperatures to kelvins

T₁ = (20 + 273.15) K = 293.15 K

T₂ = (68 + 273.15) K = 341.15 K

(b) Calculate the volume

`( 1.92 * 10^(3))/(293.15) = ( V_(2))/(341.15)\\\\6.550 = ( V_(2))/(341.15)\\\\V_(2) = 6.550 * 341.15 = 2.23 * 10^(3) \text{ L}`

The new volume of the gas is
`\boxed{2.23 * 10^(3) \text{ L}}`.