Contents
Physics Topics are often described using mathematical equations, making them precise and quantifiable.
Derive the Relationship Between Pressure, Temperature and Density of a Gas
The equation of state for an ideal gas is pV = kT. The value of the constant k depends on the mass of the gas used. Keeping pressure (p) and temperature (T) constant, if the mass of a gas is doubled, the volume is also doubled. So k is doubled. We can conclude that k is directly proportional to the mass of the gas.
When one gram-molecule or one mole of an ideal gas is taken, the constant k is written as R. As per Avogadro’s law, at the same temperature and pressure, volume of 1 gram-molecule of any gas is the same. Therefore, the value of the constant R, for all 1 gram-molecule ideal gases, is the same. Hence, it is called the universal gas constant or molar gas constant.
Hence, for 1 gram-molecule of any ideal gas,
pV = RT …….. (1) f the volume of n gram-molecule of gas is V, then volume of 1 gram-molecule of gas, = \(\frac{V}{n}\)
Then from equation (1) we get,
pᐧ\(\frac{V}{n}\) = RT or, pV = nRT …………. (2)
Obviously, for mg of a gas of molecular weight M the number of Moles, n = \(\frac{m}{M}\)
∴ pV = \(\frac{m}{M}\)RT ………… (3)
Comparing equation (3) with the gas equation pV = kT,
k = \(\frac{m}{M}\)R
Therefore for 1 gram of gas, k = \(\frac{R}{M}\) = r (say). r is also a constant and is called the specific gas constant. As the molecular weights of different gases are different, the magnitude of this constant is also different for different gases.
Magnitude of universal or molar gas constant:
From equation (1) above, the relation R = \(\frac{p V}{T}\) is valid for any ideal gas at any temperature and pressure.
Hence, R is also equal to \(\frac{p_0 V_0}{T_0}\)where p0 = 76 cmHg (standard pressure) = 76 × 13.596 × 980.665 dyn ᐧ cm-2, T0 (standard temperature) = 0 + 273 = 273 K and V0 = the volume at STP of 1 mol of a gas = 22.4 litre, as per Avogadro’s law.
Magnitude of specific gas constant: Value of r for a gas is r = \(\frac{R}{M}\).
Numerical Examples
Example 1.
Mass of 1 litre of hydrogen at STP is 0.0896 g. Cal-culate the value of R from this data.
Solution:
Volume of 0.0896 g of hydrogen at STP = 1000 cm3.
Hence, volume of 2 g or 1 mol of hydrogen at STP
= \(\frac{1000 \times 2}{0.0896}\) = 22321.4 cm3
Standard pressure = 76 × 13.6 × 981 dyn ᐧ cm-2; standard temperature = 0°C = 273 K
∴ R = \(\frac{p V}{T}\) = \(\frac{76 \times 13.6 \times 981 \times 22321.4}{273}\)erg ᐧ mol-1 ᐧ K-1
= 8.29 × 107erg ᐧ mol-1 ᐧ K-1.
Example 2.
Mass of 3.76 litre of oxygen at 2 standard atmosphere pressure and 20°C is 10 g. Find the value of R.
Solution:
Given, p = 2 standard atmosphere
= 2 × 76 × 13.6 × 981 dyn ᐧ cm-2,
V = 3.76 × 103 cm3, T = 273 + 20 = 293 K and n = \(\frac{10}{32}\) mol, [where 32 is the molecular weight of oxygen]
∴ pV = nRT
or, R = \(\frac{p V}{n T}\) = \(\frac{2 \times 76 \times 13.6 \times 981 \times 3.76 \times 10^3 \times 32}{10 \times 293}\)
= 8.33 × 107 erg ᐧ mol-1ᐧK-1
Example 3.
Density of air at STP = 1.293 g ᐧ L-1 and that of mercury = 13.6 g ᐧ cm-3. Find the value of the gas constant for 1 g of air.
Solution:
If k is the constant for 1 g of air, then k = \(\frac{1000}{1.293}\)cm3
According to the problem, 1.293 g of air occupies a volume of 1000 cm3
∴ 1 g of air occupies a volume of \(\frac{1000}{1.293}\) cm3
Here, p = 76 × 13.6 × 981 dyn ᐧ cm-1, T = 273 K.
∴ k = 76 × 13.6 × 981 × \(\frac{1000}{1.293}\) × \(\frac{1}{273}\)
= 0.287 × 107 erg ᐧ g-1 ᐧ K-1.
Example 4.
The masses, volumes and pressures of two samples of oxygen and hydrogen gases are equal. Find the ratio of their absolute temperatures.
Solution:
Let the volume of each gas = V, the pressure of each gas = p, the absolute temperatures of the gases be T1 and T2 respectively. If the mass of each sample is m g, then number of moles of the gases are respectively,
n1 = \(\frac{m}{32}\) and n2 = \(\frac{m}{2}\)
As per the equation pV = nRT,
for oxygen gas, pV = \(\frac{m}{32}\)RT1 and for hydrogen gas,
pV = \(\frac{m}{2}\)RT2
∴ \(\frac{m}{32}\)RT1 = \(\frac{m}{2}\)RT2 or, \(\frac{T_1}{T_2}\) = \(\frac{32}{2}\) = 16 : 1
Relation Between Pressure Temperature And Density of a Gas
Let an ideal gas is taken of molecular weight M (say).
For m1 mass of the gas, the volume and density be V1 and ρ1 respectively at a temperature T1K. When its mass is m2, the volume and density are V2 and ρ2 respectively at T2K.
∴ V1 = \(\frac{m_1}{\rho_1}\) and V2 = \(\frac{m_2}{\rho_2}\)
If p1 and p2 are the pressures at those temperatures, then from the equation pV = nRT = \(\frac{m}{M}\)RT we have,
\(\frac{p_1 V_1}{m_1 T_1}\) = \(\frac{p_2 V_2}{m_2 T_2}\)
[Since for a particular gas M is constant]
or, \(\frac{p_1}{\rho_1 T_1}\) = \(\frac{p_2}{\rho_2 T_2}\)
∴ \(\frac{p}{\rho T}\) = constant
i) If the pressure is a constant,
ρ1T1 = ρ2T2 i.e. ρT = constant or, ρ ∝ \(\frac{1}{T}\)
i.e., at constant pressure, the density of a gas is inversely proportional to its absolute temperature.
ii) the temperature is a constant,
\(\frac{p_1}{\rho_1}\) = \(\frac{p_2}{\rho_2}\) or, \(\frac{p}{\rho}\) = constant or,
ρ ∝ p
i.e., at a constant temperature, the density of a gas is directly proportional to its pressure.
Numerical Examples
Example 1.
Temperature and pressure on top of a hill are 7°C and 70 cmHg, and the corresponding values at its base are 27 °C and 76 cmHg. Compare the densities of air at the top and the base of the hill.
Solution:
Example 2.
Density of argon at 27 °C and 76 cmHg pressure is 1.6 g ᐧ L-1. An electric bulb of volume 200 cm3 is filled with argon. The pressure of the gas inside the bulb is 75 cmHg and the average temperature is 127°C. Find the mass of argon gas in the bulb.
Solution:
Let the density of the gas in the bulb at a pressure of 75 cmHg and at 127°C = ρ2
Here, p1 = 76 cmHg, ρ1 = 1.6 g ᐧ L-1,
T1 = 273 + 27 = 300 K
p2 = 75 cmHg, T2 = 273 + 127 = 400 K
Example 3.
At a place, air pressure is 75 cmHg and temperature is 27 °C. At another place, the respective values are 70 cmHg and 17°C. Compare the densities of air in the two places.
Solution:
We have, \(\frac{p_1}{\rho_1 T_1}\) = \(\frac{p_2}{\rho_2 T_2}\)
Given, p1 = 75 cmHg, T1 = 273 + 27 = 300 K,
p2 = 70 cmHg
and T2 = 273 + 17 = 290 K
∴ \(\frac{75}{\rho_1 \times 300}\) = \(\frac{70}{\rho_2 \times 290}\) or, \(\frac{\rho_1}{\rho_2}\) = \(\frac{75 \times 290}{300 \times 70}\) = \(\frac{29}{28}\)
∴ ρ1 : ρ2 = 29 : 28.
Example 4.
When an air bubble rises from the bottom of a lake to the upper surface, its diameter increases from 1 mm to 2 mm. If the atmospheric pressure is 76 cmHg, calculate the depth of the lake. Density of mercury is 13.6 g ᐧ cm-3.
Solution:
Let the depth of the lake be h cm.
Pressure on the air bubble, at the bottom of the lake, p1 = atmospheric pressure + pressure of water at depth h = (76 × 13.6 × 981 + h × 1 × 981) dyn ᐧ cm-2, and pressure at the surface of the lake, p2 = atmospheric pressure = 76 × 13.6 × 981 dyn ᐧ cm-2
Volumes of the bubble :
at the bottom, V1 = \(\frac{4}{3} \pi\left(\frac{1}{20}\right)^3\) cm3;
at the top, V2 = \(\frac{4}{3} \pi\left(\frac{1}{10}\right)^3\) cm3
Using Boyle’s law: p1V1 = p2V2, we get,
Example 5.
An electronic vacuum tube is constructed and sealed at 27°C and 1.2 × 10-6cmHg pressure. The tube has a volume of 100 cm3. Calculate the number of gas molecules left in the tube. Avogadro number is 6.02 × 1023, and the gas occupies a volume of 22.4 litre at STP.
Solution:
Molar density at STP
= \(\frac{6.02 \times 10^{23}}{22.4 \times 10^3}\) molecules ᐧ cm-3
Let the number of molecules left in the tube, after sealing, be n.
Example 6.
While constructing a bulb of volume 250 cm3, it is sealed at 27° C temperature and 10-3 mmHg pressure. Find the number of gas molecules in the bulb. Avogadro number = 6.0 × 1023.
Solution:
Volume of air inside the bulb, V1 = 250 cm3; pressure of air, p1 = 10-3 mmHg = 10-4 cmHg; temperature T1 = 27 + 273 = 300 K. Let at STP, the volume of air = V2 cm3; p2 = 76 cmHg; T2 = 273 K
∴ Using \(\frac{p_1 V_1}{T_1}\) = \(\frac{p_2 V_2}{T_2}\) we get
V2 = \(\frac{p_1 V_1 T_2}{T_1 p_2}\) = \(\frac{10^{-4} \times 250 \times 273}{300 \times 76}\) = 29.934 × 10-5cm3
At STP, number of molecules in air of volume 22400 cm3 = 6 × 1023.
So, that in volume 29.934 × 10-5 cm3
= \(\frac{6 \times 10^{23}}{22400}\) × 29.934 × 10-5 = 8.018 × 1015.
Example 7.
Two containers of volume 5 L and 3 L contain air at 3 standard atmospheres and 7 standard atmospheres respectively. The containers are now connected by a short narrow tube. What will be the common pressure in both the containers?
Solution:
Suppose the containers have n1 and n2 number of moles of the gas.
∴ 5 × 3 = n1RT …… (1)
and 3 × 7 = n2RT …….. (2)
Therefore, 15 + 21 = (n1 + n2)RT [adding (1) and (2)]
or, (n1 + n2)RT = 36
At constant temperature, let the common pressure attained be p.
At that common pressure p,
p(5 + 3) = (n1 + n2)RT = 36
∴ p = \(\frac{36}{8}\) = 4.5 standard atmospheres.
Example 8.
Two bulbs of equal volume are connected by a nar-row tube of negligible volume and filled with a gas at STP. If one of the bulbs is kept in melting ice and the other in a water bath at 62°C, what will be the new pressure of the gas?
Solution:
Let volume of each bulb = V cm3 and each contains n number of moles of the gas.
∴ Total number of moles contained in the bulb initially
= 2n = \(\frac{76 \times 2 V}{R \times 273}\) [from pV = nRT] ……… (1)
Let the final pressure in both the bulbs = p.
Number of moles in one of the bulbs = \(\frac{p V}{R \times 273}\)
and that in the other bulb = \(\frac{p V}{T(273+62)}\) = \(\frac{p V}{R \times 335}\)
∴ Total number of moles contained in the bulbs
= \(\frac{p V}{R \times 273}\) + \(\frac{p V}{R \times 335}\) = \(\frac{p V}{R}\left(\frac{1}{273}+\frac{1}{335}\right)\) ………. (2)
∴ From (1) and (2) we get,
\(\frac{76 \times 2 V}{R \times 273}\) = \(\frac{p V}{R}\left[\frac{1}{273}+\frac{1}{335}\right]\)
or, \(\frac{p l}{R}\)(0.037 + 0.0030) = \(\frac{76 \times V \times 2}{R \times 273}\)
or, p × 0.0067 = 0.5568 or, p = \(\frac{0.5568}{0.0066}\) = 83.10 cmHg.
Example 9.
An air bubble rises from the bottom of a lake to its upper surface. The diameters of the bubble at the bottom and the surface are 3.6 mm and 4 mm respectively. Depth of the lake is 2.5 m and the tem-perature at the upper surface is 40°C. Find the tem-perature at the bottom of the lake. Ignore the change in density of water with height. (Atmospheric pressure = 76 cmHg and g = 980cm ᐧ s-2.)
Solution:
Pressure at the bottom of the lake
p1 = 76 × 13.6 × 980 + 250 × 1 × 980
= (76 × 13.6 + 250) × 980 dyn ᐧ cm-2
Volume of the air bubble at the bottom
V1 = \(\frac{4}{3}\)π(0.18)3 cm3
Let the temperature at the bottom of the lake = T1
Now, at the surface, pressure
p2 = 76 × 13.6 × 980 dyn ᐧ cm-2;
volume of the air bubble V2 = \(\frac{4}{3}\)π(0.2)3 cm3;
temperature T2 = 273 + 40 = 313 K
Example 10.
A balloon at STP can lift a total mass of 175 kg attached with it. When the barometer reads 50 cmHg and the temperature becomes -10°C at an upper point to where the balloon rises, find the maximum mass that can be lifted. Consider the volume of the balloon to be a constant.
Solution:
The change in lifting capacity is due to the change in the upthrust, as the density of air changes at higher altitude due to the change in temperature and pressure.
Let V = volume of the balloon, Py = density at STP, p2 = den-sity at 50 cmHg and -10°C; M = mass it can carry at the given altitude.
Hence, from Archimedes’ principle,