Question

A gas has an initial volume of 52.3 L at 273 Kelvin. What is its temperature when the volume reaches 145.7 L?

Answer 1

`\boxed {\boxed {\sf 761 \ K}}`

Step-by-step explanation:

We are asked to find the new temperature of a gas after a change in volume. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

`\frac {V_1}{T_1}= (V_2)/(T_2)`

The volume is initially 52.3 liters at a temperature of 273 Kelvin.

`\frac {52.3 \ L}{273 \ K}= (V_2)/(T_2)`

The volume reaches 145.7 liters at an unknown temperature.

`\frac {52.3 \ L}{273 \ K}= (145.7 \ L )/(T_2)`

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

`52.3 \ L * T_2 = 273 \ K * 145.7 \ L`

Now the variable is being multiplied by 52.3 liters. The inverse of multiplication is division. Divide both sides by 52.3 L.

`\frac {52.3 \ L * T_2 }{52.3 \ L}=( 273 \ K * 145.7 \ L)/(52.3 \ L)`

`T_2=( 273 \ K * 145.7 \ L)/(52.3 \ L)`

The units of liters cancel.

`T_2=( 273 \ K * 145.7 )/(52.3 )`

`T_2 = 760.5372849 \ K`

The original measurements have at least 3 significant figures, so our answer must have 3. For the number we found, that is the ones place. The 5 in the tenths place tells us to round the 0 up to a 1.

`T_2 \approx 761 \ K`

When the volume reaches 145.7 liters, the temperature is 761 Kelvin.