Question

A 581.2 mL sample of carbon dioxide was heated to 347 K.

Answer 1

Answer:

The temperature at 581.2 mL volume was 242.8 K.

Explanation:

Using Charle's law  

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

Given ,  

V₁ = 581.2 mL

V₂ = 830.5 mL

T₁ = ?

T₂ = 347 K

Using above equation as:

$\frac {581.2\ mL}{T_1}=\frac {830.5\ mL}{347\ K}$

$T_1=\frac{581.2\times 347}{830.5}\ K=242.8\ K$

The temperature at 581.2 mL volume was 242.8 K.