Problems on Physical Chemistry appear really as problems to the most of students studying in different Schools, Colleges and Different universities. To Overcome this Problems to achieve a good performance in Examinations i will discuss some of these problems Topic Wise. This page is Updated Regularly so Keep on Touch With this Page.
 Problem 1:

At 1 atm pressure and 300 K temperature, the volume of the gas is 2000 cm^{3}, then calculate the volume of this gas at 600 K temperature and 2 atm pressure.
 Solution:

From the Combination of Boyle's and Charl's Law is,
(P_{1}V_{1})/T_{1} = (P_{2}V_{2})/T_{2}
Here, P_{1} = 1 atm; V_{1} = 2000 cm^{3} and T_{1} = 300K and P_{2} = 2 atm; T_{2} = 600K and V_{2} = ?
∴ 1 × 2000/300 = 2 × V_{2}/600
or, V_{2} = (1 × 2000 × 600)/(300 × 2)
= 2000 cm^{3}
 Problem 2:

Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 27^{0}C. Given, a = 1.4 atm lit^{2} mol^{2} and b=0.04 lit mol^{1} using Van der Waals equation. Also calculate the pressure of the gas using ideal gas equation and find the extent of deviation from ideal behavior.
 Problem 1:
 Solution:

For n mole Real Gas, Van der Walls equation of state is,
(P + an^{2}/V^{2})(V  nb) = nRT

In the given problem n = 2 mol, a = 1.4 atm lit^{2} mol^{2}, b = 0.04 lit mol^{1}, V = 10 lit, R = 0.082 lit atm mol^{1} K^{1} and T = (273 + 27)K = 300K.

Now, P = {nRT/(V  nb)}  (an^{2}/V^{2})
P
= {(2×0.082×300)/(102×0.04)}  (1.4×22/10)
= (5.45  0.56)
= 4.89 atm

If the gas behaves ideally its pressure, P_{ideal} = nRT/V = (2 × 0.082 × 300)/10 = 4.92 atm P_{real} ã„‘ P_{ideal} due to intermolecular attraction.
Thus the extent of deviation from ideal behavior, = (4.92  4.89) = 0.300.
 Problem 3:

Calculate the wt of O_{2} necessary to fill up a cylinder of 5 liter capacity at 0^{0}C and 100 atm. Pressure when the compressibility factor is 0.96.
 Solution:

We know that, n = g/M gm mol^{1} and Z = compressibility factor.
PV = ZnRT
or, n = PV/ZRT
∴ g/M = PV/ZRT
g = PVM/ZRT
∴ Weight of O_{2} = PVM/ZRT
= (100 × 5 × 32)/(0.96 × 0.082 × 273)
= 744.4 gm
 Problem 4:

One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors founds from the curves are 1.072 and 1.375 respectively at the initial and final states. Calculate the final volume.
 Solution:

We have from compressibility factor,
Z = PV/nRT
Thus, Z_{1} = P_{1}V_{1}/nRT_{1}
and Z_{2} = P_{2}V_{2}/nRT_{2}

Where P_{1}, V_{1} and T_{1} is the initial pressure, volume and temperature and P_{2}, V_{2} and T_{2} is the final pressure, volume and temperature.

∴ Z_{2}/Z_{1} = (P_{2}V_{2}/nRT_{2})/(P_{1}V_{1}/nRT_{1})
or, V_{2} = (Z_{2}/Z_{1})×(T_{1}/T_{2})×(P_{1}V_{1}/P_{2})
Thus the final volume,= (1.375/1.072)×(273/473)×(300 1/600)
= 370.1 c.c.
 Problem 5:

Find the Units and Dimensions of Van dar Walls constant 'a' and 'b' from Van der walls Equation.
 Solution:

For n mole Real Gas, Van der Walls equation of state is,
(P + an^{2}/V^{2})(V  nb) = nRT 

From the above equation,
Unit of an^{2}/V^{2} = Unit of Pressure.
Thus, Unit of 'a' × (mol^{2}/lit^{2}) = atm
Unit of 'a' = atm lit^{2} mol^{2}
In CGS system,unit of 'a'= dyne cm^{2} cm^{6} mol^{2}
= dyne cm^{4} mol^{2}
In SI system,unit of 'a'= Newton m^{2} m^{6} mol^{2}
= Newton m^{4} mol^{2}
In SI system,unit of 'b' = m^{3} mol^{1}
In CGS system,unit of 'b' = cm^{3} mol^{1}
∴ Dimension of 'a' = [M L^{5} T^{2} mol^{1}] and
dimension of 'b' = [L^{3} mol^{1}]
 Problem 6:

Calculate the Boyle temperature of nitrogen gas given a = 1.4 atm lit^{2} mol^{2} and b=0.04 lit mol^{1}.
 Solution:

We know that Boyle temperature,
T_{B} = a/Rb
The the Boyle temperature of Nitrogen Gas,
T_{B} = 1.4/(0.082 × 0.04) K
= 427 K
 Problem 7:

Comment on the possibility of defining a Boyle Temperature if
a = 0 and a =b = 0.
 Solution:

When, a = 0, Van der Waals Equation,
P (V  b) = RT
or, PV = RT + Pb
Differentiating with respect of pressure at constant temperature,
[d(PV)/dP]_{T} = b
But b ≠ 0 hence [d(PV)/dP]_{T} ≠ 0.

The gas has no Boyle temperature since at any temperature since at any temperature, [d(PV)/dP]_{T} ≠ 0.

Again when a = b = 0,
Van der Walls equation becomes,
PV = RT
or, [d(PV)/dP]_{T} = 0

That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.
 Problem 8:

At 273 K and under pressure of 100 atm the compressibility factor of O_{2} is 0.97. Calculate the mass of O_{2} necessary to fill a gas cylinder of 108.5 lit capacity under the given conditions.
 Solution:

From the given data, we have T = 273 K, Z = 0.97 and P = 100 atm.
From the definition of compressibility factor,
Z = PV/RT where V is the molar Volume.
Thus the molar volume of O₂ is,
V_{m} = ZRT/P
= (0.97 × 0.082 × 273 K)/(100)
= 2.17 lit mol1

The mass of this molar volume will be equal to the molar mass of oxygen, that is 2.17 lit of oxygen equal to 32 gm.Thus the mass of oxygen required to fill the gas cylinder 108.5 lit under the given condition,

= {(32 gm)/(2.17 lit)} × (108.5 lit)
= 1600 gm
= 1.6 Kg