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

Answer:

$\boxed{2.23 \times 10^{3} \text{ L}}$

Explanation:

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

$\dfrac{ V_{1} }{T_{1}} = \dfrac{ V_{2} }{T_{2}}$

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

$\dfrac{ 1.92 \times 10^{3}}{293.15} = \dfrac{ V_{2}}{341.15}\\\\6.550 = \dfrac{ V_{2}}{341.15}\\\\V_{2} = 6.550 \times 341.15 = 2.23 \times 10^{3} \text{ L}$

The new volume of the gas is $\boxed{2.23 \times 10^{3} \text{ L}}$.