Question

A gas occupies 1.00 L at standard temperature. What is the volume at 333.0°C?

Answer 1

We use Charles's Law: V1/T1=V2/T2

Standard Temperature: 0 degree Celsius= 273K
333.0 degrees Celsius= 606K

Set up: (1.00L)/ (273K)= V2/ (606.0K)
⇒ V2= (1.00L)/ (273K)* (606.0K)= 2.22L

Hope this would help :))

Answer 2

The new volume of the gas as the temperature is increased is 2.22L.

Given the data in the question;

• Initial volume of gas; $V_1 = 1.00L$

• Initial temperature; (Standard temperature); $T_1 = 273.15K$

• Final temperature; $T_2 = 333.0^oC = 606.15K$

• Final volume of gas; $V_2 = \ ?$

Charles's law

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

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

To determine the new volume as the temperature is increased, we substitute our given values into the expression above.

$\frac{V_1}{T_1} = \frac{V_2}{T_2} \\\\V_1T_2 = V_2T_1\\\\V_2 = \frac{V_1T_2}{T_1}\\ \\V_2 = \frac{1.00L\ *\ 606.15K}{273.15K}\\\\V_2 = \frac{606.15LK}{273.15K}\\\\V_2 = 2.22L$

Therefore, the new volume of the gas as the temperature is increased is 2.22L.

Learn more about Charles's law here: brainly.com/question/12835309