Question

Question #74b54

Answer 1

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, `{4.17 L}/{998 K} = V_2/{448 K}`

Multiplying both sides by 448 K, we get:

or, `{4.17 L}/{998 K}X448 K = V_2`

Rearranging,

or, `V_2={4.17 L}/{998 K}X448 K`

or, `V_2=1.87 L`

Therefore, the new volume is 1.87 L .