Question

A balloon originally had a volume of 6.5 L at 280 K the volume must be cooled to ___K to reduce volume to 3.3

Answer 1

`\boxed {\boxed {\sf 142.2 \ K}}`

Step-by-step 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:

`(V_1)/(T_1)=(V_2)/(T_2)`

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.

`\frac {6.5 \ L}{280 \ K}=(3.3 \ L)/(T_2)`

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.

`280 \ K * 3.3 \ L = 6.5 \ L * T_2`

Multiply the left side.

`924 \ K *L=6.5 \ L *T_2`

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.

`\frac { 924 \ K*L}{6.5 \ L}= ( 6.5 \ L *T_2)/( 6.5 \ L)`

`\frac { 924 \ K*L}{6.5 \ L}= T_2`

The units of liters (L) cancel.

`\frac { 924 \ K}{6.5 }= T_2`

`142.153846 \ K= T_2`

Let's round to the nearest tenth place. The 5 in the hundredth place tells us to round the 1 up to a 2.

`142.2 \ K= T_2`

The temperature must be cooled to approximately 142.2 Kelvin.