Question

Radon is a radioactive gas found below ground. If a sample of radon occupies 29 mL at 25 oC, what volume will it occupy at 500 K? Assume constant pressure.

Answer 1

Answer:

48.63 mL

Explanation:

Using Charles's law  

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,  

V₁ = 29 mL

V₂ = ?

T₁ = 25 °C

T₂ = 500 K

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

Using above equation as:

$\frac{29\ mL}{298.15\ K }=\frac{V_2}{500\ K}$

$V_2=\frac{29\cdot \:500}{298.15}\ mL$

New volume = 48.63 mL