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# Charles Law Formula

In the 1800s a French scientist named Jacques Charles made discoveries regarding the effect of temperature on gases. Charles law states that at a constant pressure the volume of a given mass of gas is directly proportional to the absolute temperature.

Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion.

Charles Law Formula relationship is expressed by

### V1/T1 = V2/T2

Where,
• V1 = original volume
• V2 = new volume
• T1 = original temperature
• T2 = new temperature
Where V and T are absolute measurements of volume and temperature at times or conditions 1 and 2.

Charles's Law Formula can be expressed by the equation.

### Vi/Ti = Vf/Tf

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## Charles Law Problems

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Solved problems based on Charles law are given below.

### Solved Examples

Question 1: Earlier we found that volume of oxygen can be obtained from a particular tank at 1.00atm and 21oC is 785L (including the volume remaining in the tank). What would be the volume of oxygen if the temperature had been 28oC?
Solution:

Ti = (21 + 273)K = 294K
Tf = (28 + 273)K = 301K

Following is the data table:

 Vi = 785L Pi = 1.00atm Ti = 294K Vf = ? Pf = 1.00atm Tf = 301K

Note that T varies and P remains constant, so V must change. These are the conditions needed to apply Charles Law.

Vf = Vi $\times$ $\frac{T_{f}}{T_{i}}$

Vf = 785L $\times$ $\frac{301K}{294K}$ = 804L

Question 2: A gas occupies a volume of 200cm3 at 0oC and 760 mm Hg. What volume will it occupy at 100oC and 760 mm Hg?
Solution:

Since the pressure is constant, there is a problem in which only the temperature is changed. We can solve by substituting in the Charles law formula, provided we convert the Celsius temperatures to absolute temperatures. Only the absolute temperatures can be used in calculations involving gases.

V1 = 221cm3; T1 = 273K (0 + 273); T2 = 373K (100 + 273)

Now we use the formula $\frac{221\ cm^{3}}{V_{2}}$ = $\frac{273K}{373K}$

V2 = $\frac{221\ cm^{3} \times 373K}{273K}$ = 302 cm3

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