Answer:
[tex]\boxed {\boxed {\sf 142.2 \ K}}[/tex]
Explanation:
This problem asks us to find the temperature change necessary to make the volume change. We use Charles's Law which states that temperature is directly proportional to the volume of a gas. The formula is:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
We know the balloon originally had a volume of 6.5 liters and a temperature of 280 Kelvin. Then, the temperature was cooled so the new volume is 3.3 liters. However, the exact new temperature is unknown.
Substitute all known values into the formula.
[tex]\frac {6.5 \ L}{280 \ K}=\frac{3.3 \ L}{T_2}[/tex]
Now, solve the new temperature (T₂). First, cross multiply. Multiply the 1st numerator by the 2nd denominator. Then, multiply the 1st denominator by the 2nd numerator.
[tex]280 \ K * 3.3 \ L = 6.5 \ L * T_2[/tex]
Multiply the left side.
[tex]924 \ K *L=6.5 \ L *T_2[/tex]
We must isolate the variable. Currently, it is being multiplied by 6.5 liters. The inverse of multiplication is division. Divide both sides by 6.5 L.
[tex]\frac { 924 \ K*L}{6.5 \ L}= \frac{ 6.5 \ L *T_2}{ 6.5 \ L}[/tex]
[tex]\frac { 924 \ K*L}{6.5 \ L}= T_2[/tex]
The units of liters (L) cancel.
[tex]\frac { 924 \ K}{6.5 }= T_2[/tex]
[tex]142.153846 \ K= T_2[/tex]
Let's round to the nearest tenth place. The 5 in the hundredth place tells us to round the 1 up to a 2.
[tex]142.2 \ K= T_2[/tex]
The temperature must be cooled to approximately 142.2 Kelvin.
