Question

We cooleda water bottle so the volume was lowered to .050 L, then we heated it to 373 K and thevolume increased to .10 L. What was the initial temperature?

Answer 1

`186.5\ \text{K}`

Step-by-step explanation:

`V_1` = Initial volume = 0.05 L

`V_2` = Final volume = 0.1 L

`T_1` = Initial temperature

`T_2` = Final temperature = 373 K

From the ideal gas law we have

`PV=nRT`

So

`V\propto T`

`(V_1)/(V_2)=(T_1)/(T_2)\\\Rightarrow T_1=(V_1)/(V_2)* T_2\\\Rightarrow T_1=(0.05)/(0.1)* 373\\\Rightarrow T_1=186.5\ \text{K}`

The intial temperature is
`186.5\ \text{K}`.