Answer:
–272.96 °C
Step-by-step explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 27.0 °C
Initial volume (V₁) = 630 L.
Final volume (V₂) = 92.0 mL
Final temperature (T₂) =?
Next, we shall convert 27.0 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 27.0 °C
Initial temperature (T₁) = 27.0 °C + 273
Initial temperature (T₁) = 300 K
Next, we shall convert 92.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
92 mL = 92 mL × 1 L / 1000 mL
92 mL = 0.092 L
Next, we shall determine the final temperature.
Initial temperature (T₁) = 300 K
Initial volume (V₁) = 630 L.
Final volume (V₂) = 0.092 L
Final temperature (T₂) =?
V₁ / T₁ = V₂ / T₂
630 / 300 = 0.092 / T₂
2.1 = 0.092 / T₂
Cross multiply
2.1 × T₂ = 0.092
Divide both side by 2.1
T₂ = 0.092 / 2.1
T₂ = 0.04 K
Finally, we shall convert 0.04 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
Final temperature (T₂) = 0.04 K
Final temperature (T₂) = 0.04 – 273
Final temperature (T₂) = –272.96 °C