Given:
[tex]V_{1} = \text{6.00 L}[/tex]
[tex]T_{1} = \text{20.0°C + 273.15 = 293.15 K}[/tex]
[tex]V_{2} = \text{4.00 L}[/tex]
Unknown:
[tex]∆T[/tex]
Solution:
Step 1: Calculate the final temperature.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}[/tex]
[tex]T_{2} = T_{1} × \frac{V_{2}}{V_{1}}[/tex]
[tex]T_{2} = \text{293.15 K} × \frac{\text{4.00 L}}{\text{6.00 L}}[/tex]
[tex]T_{2} = \text{195.433 K} [/tex]
Step 2: Calculate the decrease in temperature.
[tex]∆T = T_{2} - T_{1}[/tex]
[tex]∆T = \text{195.433 K} - \text{293.15 K}[/tex]
[tex]∆T = -\text{97.72 K}[/tex]
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