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

Answer:

$186.5\ \text{K}$

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

$V\propto T$

$\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\Rightarrow T_1=\dfrac{V_1}{V_2}\times T_2\\\Rightarrow T_1=\dfrac{0.05}{0.1}\times 373\\\Rightarrow T_1=186.5\ \text{K}$

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