Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What is CP and CV of Gases?
Let masses m1, m2, … of some gases, which do not react chemically, form a mixture.
The respective specific heats at constant volume are \(c_{v_1}\), \(c_{v_2}\), …… . So, in the mixture,
heat capacity of the 1st gas = m1\(c_{\nu_1}\),
heat capacity of the 2nd gas = m2\(c_{\nu_2}\) and so on.
The total heat capacity of the mixture = m1\(c_{v_1}\) + m2\(c_{v_2}\) + ……….
As the total mass = m1 + m2 + ……, the effective specific heat of the mixture, i.e., the heat capacity per unit mass at constant volume is
cv = \(\frac{m_1 c_{v_1}+m_2 c_{v_2}+\cdots}{m_1+m_2+\cdots}\) ……… (1)
Similarly, the effective specific heat of the mixture at constant pressure is
cp = \(\frac{m_1 c_{p_1}+m_2 c_{p_2}+\cdots}{m_1+m_2+\cdots}\) ….. (2)
∴ γ = \(\frac{m_1 C_{p_1}+m_2 C_{p_2}+\ldots}{m_1 C_{v_1}+m_2 C_{v_2}+\ldots}\)
Similar expressions are obtained for the molar specific heats of a mixture of gases (see the chapter Kinetic Theory of Gases). As an example, we may consider air as a mixture of oxygen, nitrogen, water vapour, carbon dioxide, etc. But the presence of all the gases other than oxygen and nitrogen is extremely small in comparison. To calculate specific heat, we can easily ignore those gases. In air, oxygen and nitrogen behave as ideal gases and both of them are diatomic.
Now, the value of molar specific heat at constant volume (Cv) is the same for all ideal diatomic gases. So, Cv for oxygen is equal to the Cv for nitrogen. As a result, air will have the same effective value of Cv. Similar arguments are applicable for the effective molar specific heat Cp of air at constant pressure. Accordingly, the value of γ for air is the same as that of oxygen or nitrogen. But the effective specific heat of air per unit mass cannot be obtained in such a simple way. The values of cv and cp of oxygen gas are different from those of nitrogen gas. So, equations (1) and (2) have to be used. In those equations, we may put m1 = 22 g for oxygen and m2 = 78 g for nitrogen, because approximately 22% of oxygen by mass mixes with 78% of nitrogen by mass to form air.
Numerical Examples
Example 1.
The specific heat of oxygen gas at constant volume is 0.155 cal ᐧ g-1. What is its specific heat at constant pressure? Given, the molecular mass of oxygen = 32 and R = 2 cal ᐧ mol-1 ᐧ °C-1.
Solution:
Molar specific heats of oxygen gas at constant volume and constant pressure respectively, are
Cv = Mcv and Cp = Mcp
Now, Cp – Cv = R or, Cp = Cv + R or, Mcp = Mcv + R
or, cp = cv + \(\frac{R}{M}\) = 0.155 + \(\frac{2}{32}\) = 0.2175 cal ᐧ g-1 ᐧ °C-1.
Example 2.
For 1 mol of a triatomic ideal gas Cv = 3R (R is universal constant). Find δ (=Cp/Cv) for that gas. [WBCHSE Sample Question ’13]
Solution:
For 1 mol of an ideal gas, Cp – Cv = R
∴ Cp = Cv + R = 3R + R = 4R ∴ δ = \(\frac{C_p}{C_v}\) = \(\frac{4 R}{3 R}\) = \(\frac{4}{3}\).
Example 3.
The temperature of 20g of oxygen gas is raised from 10°C to 90°C. Find out the heat supplied, the rise in internal energy and the work done by the gas, if the temperature rises at
(i) constant volume,
(ii) constant pressure, Given the specific heats of oxygen are 0.155 cal ᐧ g-1 ᐧ °C-1 at constant volume, and 0.218 cal ᐧ g-1 ᐧ °C-1 at constant pressure.
Solution:
i) Heat supplied at constant volume is
Qv = mcv(tf – ti)
= 20 × 0.155 × (90 – 10) = 248 cal
As the volume is constant, the work done, W = 0
From the first law of thermodynamics, rise in internal energy is Uf – Ui = Q – W = 248 – 0 = 248 cal.
ii) Heat supplied at constant pressure is
Qp = mcp(tf – ti)
= 20 × 0.218 × (90 – 10) = 348.8 cal
Now, oxygen may be considered as an ideal gas. The temperature rise is the same in both cases. As internal energy is a function of temperature only, the rise in internal energy will also be the same.
∴ Uf – Ui = 248 cal
Then, Uf – Ui = Q – W
or, W = Q – (Uf – Ui) = 348.8 – 248 = 100.8 cal.
Example 4.
Find out the molar specific heats Cp and Cv of an ideal gas having γ = 1.67. Given, R = 2 cal ᐧ mol-1 ᐧ °C-1.
Solution:
γ = \(\frac{C_p}{C_v}\) or, Cp = γCv
For an ideal gas,
Cp – Cv = R, γCv – Cv = R or, Cv(γ – 1) = R
or, Cv = \(\frac{R}{\gamma-1}\) = \(\frac{2}{1.67-1}\) = \(\frac{2}{0.67}\) = 2.985 cal ᐧ mol-1 ᐧ °C-1.
∴ Cp = Cv + R = 2.985 + 2 = 4.985 cal ᐧ mol-1 ᐧ °C-1
Example 5.
A gas of density 0.00125 g ᐧ cm-3, volume 8 L at 0°C temperature and 1 atm pressure is supplied with 30 cal of heat to raise its temperature to 15°C at constant pressure. Determine the specific heat of the gas at constant pressure and at constant volume. Given, R = 2 cal ᐧ °C-1.
Solution:
Mass of the gas,
m = (8 × 103 cm3) × 0.00125 g ᐧ cm-3 = 10 g.
Heat gained by the gas at constant pressure,
Q = mcp(tf – ti)
or, cp = \(\frac{Q}{m\left(t_f-t_i\right)}\) = \(\frac{30}{10 \times(15-0)}\) = 0.2 cal ᐧ g-1 ᐧ °C-1
Now, mass of 8 L of the gas at STP = 10 g.
∴ Mass of 22.4 L of the gas at STP = \(\frac{10 \times 22.4}{8}\) = 28 g.
∴ Molecular mass of the gas, M = 28.
So, molar specific heat at constant pressure,
Cp = Mcp = 28 × 0.2 cal ᐧ mol-1 ᐧ °C-1
Molar specific heat at constant volume,
Cv = cp – R = (28 × 0.2 – 2) cal ᐧ mol-1 ᐧ °C’
∴ Specific heat of the gas at constant volume,
cv = \(\frac{C_v}{M}\) = \(\frac{28 \times 0.2-2}{28}\) = 0.2 – \(\frac{2}{28}\)
= 0.1286 cal ᐧ g-1 ᐧ °C-1.
Example 6.
The temperature of 20g of oxygen gas is raised from 50°C to 100°C
(i) at constant volume and
(ii) at constant pressure.
Find out the amount of heat supplied and the rise in internal energy in each case.
Given, R = 2 cal ᐧ mol-1. K-1 ; for oxygen, cv = 0.155 cal ᐧ g-1 . °C-1. [WBJEE ’05]
Solution:
i) Heat supplied at constant volume and
Qv = mcv(tf – ti)
= 20 × 0.155 × (100 – 50) = 155 cal
Work done, W = 0
So, rise in internal energy,
Uf – Ui = Q – W = 155 – 0 = 155 cal
= 155 × 4.2 J = 651 J.
ii) Oxygen may be taken as an ideal gas. Internal energy is a function of temperature only.
So, the rise in internal energy at constant pressure will be the same, i.e., Uf – Ui = 155 cal = 651 J.
Cv = Mcv = 0.155 × 32 = 4.96 cal ᐧ mol-1 ᐧ K-1
[M = molecular mass of oxygen = 32]
∴ Cp = Cv + R = 4.96 + 2 = 6.96 cal ᐧ mol-1 ᐧ K-1
20 g oxygen = \(\frac{20}{32}\) mol oxygen
∴ Heat supplied at constant pressure,
= \(\frac{20}{32}\) × 6.96 × (100 – 50)
= 217.5 cal.
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