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A gas occupies 12.3 L at a temperature of 40.0 K. What is the volume when the temperature is increased to 60.0 K?

Sagot :

Answer:

the volume is 18.45 L

Explanation:

The computation of the volume when the temperature is increased to 60.0 K is shown below:

Since the gas occupied 12.3 L at a temperature of 40.0 k

And, the volume when the new temperature is 60.0 k

So, the volume is

= 60.0 k × 12.3 L ÷ 40.0 k

= 18.45 L

hence,  the volume is 18.45 L