Answer:
5542.37 K
Explanation:
Given that,
Initial volume, V₁ = 590 mL
Initial temperature, T₁ = -55°C = -55 + 273 = 218 K
Final volume, V₂ = 15 L
We need to find the final 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}[/tex]
Where
T₂ is final volume of the gas
Put all the values in above relation,
[tex]T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{15\ L\times 218\ K}{590\times 10^{-3}\ L}\\\\T_2=5542.37\ K[/tex]
So, the required temperature is equal to 5542.37 K.
