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?

Question

asked Jun 8, 2021 40.1k views

1 Answer

Artem Sobolev · Answer 1 · 2021-06-13T10:06:55+0000

Answer:

$186.5\ \text{K}$

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
$186.5\ \text{K}$ .