# Question #74b54

Mar 28, 2017

The new volume of the ethane gas is 1.87 L.

#### Explanation:

In order to solve this question, we need to use Charles's law which states that the volume of a gas is proportional to its temperature in Kelvin. Mathematically,

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

Where ${V}_{1}$ is the initial volume of the gas, ${T}_{1}$ is the initial temperature of the gas (in Kelvin), ${V}_{2}$ is the volume of the gas after heating (or cooling, if that's what you did), and ${T}_{2}$ is the final temperature of the gas (in Kelvin).

Note: Please remember that the temperature is in Kelvin scale not in Degree scale.

Now, from the question we have,

Initial Volume of Ethane gas (${V}_{1}$) = 4.17 L
Initial Temperature of Ethane gas (${T}_{1}$) = 725 + 273 = 998 K
Final Volume of Ethane gas (${V}_{2}$) = ?
Final Temperature of Ethane gas (${T}_{2}$) = 175 + 273 = 448 K

Using the Charle's law:

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

Plug the given values in above laws, we get:

or, $\frac{4.17 L}{998 K} = {V}_{2} / \left\{448 K\right\}$

Multiplying both sides by 448 K, we get:

or, $\frac{4.17 L}{998 K} X 448 K = {V}_{2}$

Rearranging,

or, ${V}_{2} = \frac{4.17 L}{998 K} X 448 K$

or, ${V}_{2} = 1.87 L$

Therefore, the new volume is 1.87 L .