Answer:
[tex]\boxed{2.23 \times 10^{3} \text{ L}}[/tex]
Explanation:
The pressure is constant, so we can use Charles' Law.
[tex]\dfrac{ V_{1} }{T_{1}} = \dfrac{ V_{2} }{T_{2}}[/tex]
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
[tex]\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}[/tex]
The new volume of the gas is [tex]\boxed{2.23 \times 10^{3} \text{ L}}[/tex].
