Answer:
670.45 K
Explanation:
Given that,
Initial volume, V₁ = 1.22 L
Initial temperature, T₁ = 286 K
Final volume, V₂ = 2.86 L
We need to find the new temperature of the gas. The relation between the volume and the temperature is given by :
[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{2.86\times 286}{1.22}\\\\T_2=670.45\ K[/tex]
So, the new temperature of the gas is equal to 670.45 K.
