Question

A gas is enclosed in a vessel at standard temperature. Find out the temperature at which the volume of enclosed gas will be 1/6 of its initial volume. (Given that the pressure remains constant).

Answer 1

Step 1- Write down the given terms:

Suppose that the initial volume of gas, V₁=x

Final Volume, V₂= x/6

Initial temperature, T₁ =0°C = 0+273= 273K

Final temperature, T₂=?

Step 2- Write the formula:

Applying Charle’s law,

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

Step 3- Put the given values:

$\frac{x}{273}$=$\frac{x}{6\times T_2}$

T₂=$\frac{273x}{6\times x}$= 45.5K

Hence, the temperature is 45.5K