## CHARLES LAW

**CHARLES LAW:**According to this law, the volume of a sample of a gas is directly proportional to its absolute Temperature (in K) at constant pressure. On increasing the Temperature the kinetic energy of the molecules increases, so they collide with the walls with greater force and because we are keeping the pressure constant this greater force results in increase of volume of the sample.

**V**α

**T (K)**

V/T = constant V = kT

V1/T1 = V2/T2 = V3/T3 = ....so on

V/T = constant V = kT

V1/T1 = V2/T2 = V3/T3 = ....so on

The variation of the volume of a
fixed amount of gas with the temperature at constant pressure. Note that in
each case the isobars extrapolate to zero volume at T = 0, or q = –273°C.

The V-T graph obtained is a
straight line passing through origin as it represents the equation y = mx where
m is the slope of the line. At 0 K the volume also becomes almost zero. The
graph is extrapolated/dotted because this condition is unattainable. However,
when the same graph is drawn with Temperature in

^{0}C then the curve shifts and it intersects the x-axis at -273.15^{0}C as shown. These graphs are called**“isobars”(constant pressure curves).****We can see
that the volume of the gas at – 273.15 °C ( 0 K) will be zero. **

**This means that gas
will not exist. In fact all the gases get liquefied**

**before this temperature is
reached. The lowest hypothetical or **

**imaginary temperature at which gases are
supposed to occupy**

**zero volume is called Absolute zero. All gases obey Charles’
law **

**at very low pressures and high temperatures.**

Q. A sample of gas at 1.20 atm and
27°C is heated at constant pressure to 57°C. Its final volume is 4.75 L. What
was its original volume?

Solution.

P1 = P2 = 1.2 atm

V

_{1}=? V_{2}= 4.75L
T

_{1}= 300K T_{2}= 330 K
From Charles’ law:

^{V}__1__

_{=}

^{V}2
T

_{1}T_{2}
V

_{1 }= 4.32L