-
States Of Matter
GAS LAWS
The gas laws are a set of laws that describe the relationship between thermodynamic temperature (T), pressure (P) and volume (V) of gases.
(I) Boyle's Law
It states "at a constant temperature (T), the pressure (P) of a given mass (or moles, n) of any gas varies inversely with the volume (V)". Pressure (P) and volume (V) of gases.
i.e. P \( \propto \frac{1} {V}\) (for given n and T)
or, PV = K = constant
For two or more gases at constant temperature
P1V1 = P2V2 = ..... = K
Also, density d \( \propto \frac{1} {V}\)
Hence P \( \propto \) d
or \( \frac{{P_1 }} {{d_1 }} = \frac{{P_2 }} {{d_2 }}\) = ......... = K -
States Of Matter
* Graphical Representation
These plots drawn at constant temperature for a gas are called Isotherms. -
States Of Matter
SOLVED EXAMPLES
Ex. A 10 cm volume of air is trapped by a column of Hg, 8 cm long in capillary tube horizontally fixed as shown below at 1 atm pressure. Calculate the length of air column when the tube is fixed at same temperature.
(a) Vertically with open end up
(b) Vertically with open end down
(c) At 45ºC from with open end up
Sol. (a) When the capillary tube is held as vertically open end up (fig.2),
The pressure on the air column = atmospheric pressure + pressure of 8 cm Hg column
= 76 + 8 = 84 cm of Hg.
Let, at this condition the length of the air column = l2 and the length of air column when capillary is horizontally fixed = l1 = 10 cm and pressure on air column = 1atm.
Let the cross section of the capillary = a cm2
\(\therefore\) 76 × 10 × a = 84 × l2 × a
or l2 = \( \frac{{76 \times 10}} {{84}}\) = 9.04 cm
(b) When the capillary tube is held as vertically open end down (fig.3), the pressure on the air column = atmospheric pressure – pressure of 8 cm Hg column
= 76 – 8 = 68 cm of Hg.
Let at this condition the length of air column = l3.
\(\therefore\) 68 × l3 × a = 76 × 10 × a
or l3 = \( \frac{{76 \times 10}} {{68}}\) = 11.17 cm
(c) When the capillary is held at 45º with open end up, the weight of Hg is partially borne by, the gas and partially by the Hg. The pressure on the gas due to Hg column
= 8 × cos 45º
= 8 ×\( \frac{1} {{\sqrt 2 }}\) = \( \frac{8} {{\sqrt 2 }}\) = \( \left( {76 + \frac{8} {{\sqrt 2 }}} \right)\) cm of Hg
\(\therefore\) total pressure on the gas = cm of Hg.
Let length of air column at this pressure = l4.
\(\therefore\) l4 × a × \( \left( {76 + \frac{8} {{\sqrt 2 }}} \right)\)= 10 × a × 76
Ex. A rubber balloon contains some solid marbles each of volume 10 ml. A gas is filled in the balloon at a pressure of 2 atm and the total volume of the balloon is 1 litre in this condition. If the external pressure is increased to 4atm the volume of Balloon becomes 625 ml. Find the number of marbles present in the balloon.
Sol. Let the no. of marbles be = n .
volume of marble = 10 n ml.
volume of balloon earlier = 1000 ml.
later = 625 ml.
Now for the gas inside the balloon temperature and amount of the gas is constant, hence boyles law can be applied
P1V1 = P2V2
4× (625 – 10n) = 2 × (1000 – 10n)
625 × 4 = 2000 – 20n + 40n
625 × 4 – 2000 = 20n
\(\frac{{625 \times 4--2000}} {{20}} \) = n. \(\frac{{125}} {5} \) = n ; n = 25