Question

A balloon containing methane gas has a volume of 2.64 L at 31.0 C. What volume will the balloon occupy at 62.0 C?

Answer 1

Answer:

The balloon will occupy the volume of 2.91 L at 62.0 °C .

Explanation:

Using Charle's law  

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

Given ,  

V₁ = 2.64 L

V₂ = ?

T₁ = 31.0 °C

T₂ = 62.0 °C  

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (31.0 + 273.15) K = 304.15 K  

T₂ = (62 + 273.15) K = 335.15 K  

Using above equation as:

$\frac{2.64}{304.15}=\frac{V_2}{335.15}$

$V_2=\frac{2.64\cdot \:335.15}{304.15}$

New volume = 2.91 L

The balloon will occupy the volume of 2.91 L at 62.0 °C .