HBSE Class 11 Physics Notes For First Law of Thermodynamics: Closed Systems

Thermodynamics – First And Second Law Of Thermodynamics Specific Heat Of A Gas

From calorimetry, we know that H = mst or s = \({H}{mt}\)

i.e., specific heat = \(\frac{m t}{\text { mass } \times \text { change in temperature }}\)

For a body of unit mass, specific heat (s)

= \(\frac{\text { heat transfer }(H)}{\text { change in temperature }(t)}\)

This definition of specific heat is useful for solids and liquids but is incomplete in case of gases. For example, if a gas is suddenly compressed, its temperature rises, even when no heat is exchanged with the surroundings. Here, H = 0, but t ≠ 0.

So, s = \(\frac{H}{t}=\frac{0}{t}=0\)

Again, if the die temperature of an expanding gas is to be kept constant, some heat must be supplied from outside. Here, H = 0, but t ≠ 0.

i.e., when change in temperature = 0, specific heat ∞

This shows that the specific heat of a gas can have any value from zero to infinity. So, the specific heat can have a definite and useful value only when some condition is imposed on the gas.

In general, when a gas is healed, not only does its temperature change, but the volume and the pressure also change at the same time.

First Law Of Thermodynamics For Closed Systems Class 11 HBSE Notes

Now, two special cases are considered:

1. The absorbed heat increases the temperature and pressure of the gas while the volume Is kept constant.
2. CTD Hie absorbed heat increases the temperature and volume of the gas while the pressure is kept constant.

Heat required by unit mass of gas to raise its temperature by one degree, keeping its volume or pressure constant, is called specific heat of the gas. So, we get definitions of two specific heal in case of a gas

1. Specific heat at constant volume and
2. Specific heat at constant pressure.

Specific heat at constant volume: Heat absorbed by unit mass of a gas to raise its temperature by one degree, keeping the volume constant, is called the specific beat of that gas at constant volume (c_v).

So, heat taken by a gas of mass m for rise in temperature t, at constant volume, is Q_v = mc_vt….(1)

Specific heat at constant pressure: Heat absorbed by the unit mass of a gas to raise its temperature by one degree, keeping the pressure constant, is called the specific heat of that gas at constant pressure (c_p).

So, heat taken by a gas of mass m for the rise in temperature t, at constant pressure,  Q_p = mc_pt…(2)

First Law Of Thermodynamics Closed System Derivation Class 11 HBSE

Heat Capacity Formula: The heat capacity of a body is defined as the heat absorbed by the body per unit rise in temperature. The above discussions show that appropriate conditions are to be imposed on the definition of heat capacity.

So, heat capacity at constant volume, \(C_v =\frac{\text { heat absorbed }}{\text { change in temperature }}=\frac{Q_v}{t}=m c_v\) [using 1]

or, heat capacity at constant volume = mass x specific heat at constant volume

Similarly, C_p = mc_p

or, heat capacity at constant pressure = mass x specific heat at constant pressure

Specific heat is an intrinsic thermodynamic property. But the relations C_v = mc_v and C_p = mc_p show that C_v and C_p are proportional to the mass. So, heat capacity is an extrinsic thermodynamic property.

Molar heat capacity or molar specific heat: if M is the molecular weight of a gas (or of any other substance) mass of 1 mol = M g. The heat capacity of 1 mol of a substance is called molar heat capacity or molar specific heat.

Molar specific heat at constant volume, C_v = Mc_v, and molar specific heat at constant pressure, C_p = Mc_p

The molecular weight M is constant for a particular substance. So, like c_v and c_p, the molar specific heats C_v and C_p are also intrinsic thermodynamic properties.

First Law Of Thermodynamics For Closed Systems Formula Class 11

Units of specific heat:

\(\begin{array}{|c|c|c|}
\hline \text { specific heat }\left(c_\nu \text { or } c_p\right) \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1} \\
\hline \begin{array}{c}
\text { molar specific heat }\left(C_\nu\right. \\
\text { or } \left.C_n\right)
\end{array} \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \mathrm{~J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \\
\hline
\end{array}\)

 

Specific heat Of solids and liquids: When a solid or from a liquid is heated, thermal expansion of volume takes place. However, a change in pressure due to a change in temperature is not usually observed in practice.

So, for solids and liquids, the specific heat at constant volume (c_v) is not a useful property. The specific heat of a solid or a liquid, as discussed in calorimetry, is actually its specific heat at constant pressure (c_p).

The observation of c_v may be very difficult, but still every solid and liquid has definite values of both c_v and c_p, just like that of a gas. For water, the specific heat at constant pressure is c_p = 1 cal · g^-1 · °C^-1 (CGS)

= 4200 J ·  kg^-1 · K^-1 (SI)

As M = 18 for water, the molar specific heat at constant pressure is,

C_p = 18 cal · mol^-1 °C^-1 (CGS)

= 18 x 4.2 J · mol^-1 K^-1 (SI)

Applications Of First Law Of Thermodynamics In Closed Systems Class 11

Relation with internal energy: The first law of thermodynamics states that dQ = dU+ dW or, dQ = dU+ pdV…(3)

If we take a process at constant volume, the heat absorbed by a mass m for a temperature change dT may be expressed as dQ_v = mc_vdT.

Here, as V = constant, dV = 0. So from equation (3), we get, mc_vdT = dU or, dU = mc_vdT…(4)

When m, c_v, and dT are known, equation (4) gives a measure of the change of internal energy.

In integral form, U_f – U_i = mc_v (T_f – T_i). It is also clear that dU or U_f– U_i, is proportional to mass. So, die difference in internal energy, and the internal energy Itself, is an extrinsic thermodynamic property.

c_p is greater than c_v: The first law of thermodynamics gives, dQ= dU+dW= dU+pdV

For a process at constant volume, dQ_v = mc_vdT and dV= 0. So, mc_vdT= dU or, dU = mc_vdT.

Now let us consider an ideal gas as the working substance. For a process at constant pressure with the same temperature change dT, we have dQ_p = mc_pdT.

As the internal energy U of a gas depends on temperature only, the value of dU in this process will be the same, i.e., dU = mc_vdT. So from the relation dQ = dU+ dW, we have for the process at constant pressure,

dQ_p = mc_vdT+ dW

or, mc_pdT = mc_vdT+ pdV or, m(c_p– c_v)dT = pdV

or, \(c_p-c_\nu=\frac{p}{m} \frac{d V}{d T}\)…..(5)

As the volume of gas increases with temperature rise, \(\frac{d V}{d T}\)>0

∴ Equation (5) shows that c_p > c_v for a gas.

In general, this is true for solids and liquids also. For any substance, the specific heat at constant pressure is greater than the specific heat at constant volume. This is because at constant volume, no work is done. So the heat absorbed changes the internal energy only.

But at constant pressure, the heat absorbed changes the internal energy and also does some work. Thus, a greater amount of heat is absorbed in the latter case, and as a result, the specific heat at constant pressure is higher.

First Law Of Thermodynamics Closed System Equation Class 11 HBSE

Difference between the Two specific heats of an ideal gas: Let us consider n mol of an ideal gas. Its volume V, pressure p, and absolute temperature T are related by the equation of state pV = nRT ….(1)

where, R = universal gas constant or molar gas

= 8.31 x 10⁷ erg · mol^-1 · °C^-1

= 8.31 J · mol^-1 · K^-1 ≈ 2 cal · mol^-1 · K^-1

From the first law of thermodynamics, dQ = dU+dW = dU+pdV…(2)

Now, suppose the temperature of the gas is raised by dT at constant volume. So, the heat taken is dQ_v = nC_vdT, where C_v = molar specific heat of the gas at constant volume.

Also, as V = constant, dV = 0. So, from equation (2), dU = nC_vdT…(3)

For an ideal gas, U is a function of T only. So, for the same change dT of temperature, dU will remain the same for all processes. Then equation (2) can be written as, dQ = nC_vdT+pdV….(4)

This equation is applicable to all processes. Now, Suppose the temperature of the gas is raised by the same amount dT at constant pressure. The heat taken is dQ_p = nC_pdT, where C_p = molar specific heat at constant pressure. So, from equation (4),

nC_pdT = nC_vdT+pdV or, n(C_p-C_v)dT= pdV

or, \(n\left(C_p-C_v\right)=p \frac{d V}{d T}\)

From (1), \(V=\frac{n R T}{p}\)

So at constant pressure, \(\frac{d V}{d T}=\frac{n R}{p} or, p \frac{d V}{d T}=n R\) Then we have, \(C_p-C_v=R\)….(5)

So, the difference between the two molar-specific heats of an ideal gas is equal to the universal gas constant. As R is a constant, C_p – C_v has the same value for all gases.

The specific heats per unit mass of an ideal gas are, \(c_v=\frac{C_v}{M} \text { and } c_p=\frac{C_p}{M} \text {, }\)

where M = molecular weight of the gas

Then, \(c_p-c_v=\frac{1}{M}\left(C_p-C_\nu\right)=\frac{R}{M}=r .\)

where, r = gas constant for the unit mass of the gas.

As the molecular weight M is different for different gases, the value of (c_p-c_v) is also different. Because of this, 1 mol of a gas is more useful in thermodynamic discussions than unit mass of a gas.

It should be noted that, in the application of equation (5), all three quantities are to be converted into the same unit. If C_p and C_v are expressed in unit of heat (cal · mol^-1 · °C^-1) and R in unit of work (erg · mol^-1 · °C^-1), then R is to be divided by J to express it in unit of heat. In this case, equation (5) can be written as, \(C_p-C_v=\frac{R}{J}\)…(6)

Examples Of First Law Of Thermodynamics In Closed System Class 11

Importance of the Ratio of the Two Molar Specific Heats: The ratio between Cp and Cv is usually denoted by the Greek letter γ:

⇒ \(\gamma=\frac{C_p}{C_v}=\frac{\text { molar specific heat at constant pressure }}{\text { molar specific heat at constant volume }}\)

As C_p> C_v, we have γ > 1. The ratio between the specific heats per unit mass is also the same because

⇒ \(\frac{c_p}{c_v}=\frac{C_p / M}{C_v / M}=\frac{C_p}{C_v}=\gamma,\) where M = molecular weight.

The value of γ is important for different applications in physics and chemistry :

1. The value of γ gives an idea of the molecular structure of any gas.

• For monatomic gases, \(\gamma=\frac{5}{3}\) = 1.67 (helium, neon, argon, etc.)
• For diatomic gases, \(\gamma=\frac{5}{3}\) = 1.4 (hydrogen, oxygen, nitrogen, carbon monoxide, etc.)
• For triatomic gases, \(\gamma=\frac{5}{3}\) = 1.33 (water vapor, carbon dioxide, ozone, etc.)
• For polyatomic gases like ammonia, methane, and many organic gases, the value of γ lies between 1.1 and 1.3. So, the knowledge of γ gives an idea about the number of atoms in a molecule of a gas.

2. The thermodynamic relationships between different properties of a gas in an adiabatic process involve the value of γ.

3. The expression for the velocity of sound in involves the value of γ.

Specific Heat of a Mixture 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 \(c_{\nu_1}\), \(c_{\nu_2}\)……

So, in the mixture, heat capacity of the 1st gas = \(m_1 c_{\nu_{1}}\), heat capacity of the 2nd gas = \(m_2 c_{\nu_{2}}\) and so on.

The total heat capacity of the mixture = \(m_1 c_{v_1}+m_2 c_{v_2}+\cdots\)…..

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

⇒ \(c_v=\frac{m_1 c_{v_1}+m_2 c_{v_2}+\cdots}{m_1+m_2+\cdots}\)….(1)

Similarly, the effective specific heat of the mixture at constant pressure is

⇒ \(c_p =\frac{m_1 c_{p_1}+m_2 c_{p_2}+\cdots}{m_1+m_2+\cdots}\)…(2)

∴ \(\gamma =\frac{m_1 C_{p_1}+m_2 C_{p_2}+\cdots}{m_1 C_{v_1}+m_2 C_{v_2}+\ldots}\)

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 vapor, carbon dioxide, etc. However, 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 (C_v) is the same for all ideal diatomic gases. So, C_v for oxygen is equal to the C_v for nitrogen. As a result, air will have the same effective value of C_v.
• Similar arguments are applicable for the effective molar-specific heat C of air at constant pressure. Accordingly, the value of γ for air is the same as that of oxygen or nitrogen. However, the effective specific heat of air per unit mass cannot be obtained in such a simple way.

The values of c_v and c_p 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.

Relation Between Internal Energy And First Law Of Thermodynamics Closed System

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Specific Heat Of A Gas Numerical examples

Example 1. The specific heat of oxygen gas at constant volume is 0.155 cal · g^-1 · °C^-1. What is its specific heat at constant pressure? Given, the molecular mass of oxygen = 32 and R = 2 cal · mol^-1 • °C^-1.
Solution:

Given

The specific heat of oxygen gas at constant volume is 0.155 cal · g^-1 · °C^-1.

Molar specific heats of oxygen gas at constant volume and constant pressure respectively, are \(C_v=M c_v \text { and } C_p=M c_p\)

Now, \(C_p-C_\nu=R or, C_p=C_\nu+R or, M c_p=M c_\nu+R\)

or, \(c_p=c_v+\frac{R}{M}=0.155+\frac{2}{32}=0.2175 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\).

Example 2. For 1 mol of a triatomic ideal gas C_v = 3R (R is the universal gas constant). Find δ (= C_p/C_v) for that gas.
Solution:

For 1 mol of an ideal gas, C_p -C_v= R

∴ \(C_p=C_v+R=3 R+R=4 R\)

∴ \(\delta=\frac{C_p}{C_v}=\frac{4 R}{3 R}=\frac{4}{3} .\)

Example 3. The temperature of 20 g 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

1. Constant volume,
2. Constant pressure. Given, the specific heats of oxygen are 0.155 cal ·  g^-1 · °C^-1 at constant volume, and 0.218 cal · g^-1 · °C^-1 at constant pressure.

Solution:

1. Heat supplied at constant volume is \(Q_v=m c_v\left(t_f-t_i\right)\)

= 20 x 0.155 x (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 U_f – U_i = Q-W= 248-0 = 248 cal.

2. Heat supplied at constant pressure is \(Q_p=m c_p\left(t_f-t_i\right)\)

= 20 x 0.218 x (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.

∴ \(U_f-U_i=248 \mathrm{cal}\)

Then, \(U_f-U_i=Q-W\)

or, \(W=Q-\left(U_f-U_i\right)=348.8-248=100.8 \mathrm{cal}\) .

Limitations Of First Law Of Thermodynamics In Closed System Class 11

Example 4. Find out the molar-specific heats C_p and C_v of an ideal gas having γ = 1.67. Given, R = 2 cal · mol^-1 · °C^-1.
Solution:

γ = \(\frac{C_p}{C_\nu} \text { or, } C_p=\gamma C_\nu\)

For an ideal gas, \(C_p-C_v=R \text { or, } \gamma C_\nu-C_v=R \text { or, } C_\nu(\gamma-1)=R\)

or, \(C_v=\frac{R}{\gamma-1}=\frac{2}{1.67-1}=\frac{2}{0.67}=2.985 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

∴ \(C_p=C_v+R=2.985+2=4.985 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} .\)

Example 5. A gas of density 0.00125 g · cm^-3, volume 8 1, at 0‘C temperature and 1 atm pressure is supplied with 50 ral 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 · mol^-1 C^-1.
Solution:

Given

A gas of density 0.00125 g · cm^-3, volume 8 1, at 0‘C temperature and 1 atm pressure is supplied with 50 ral of heat to raise its temperature to 15 C at constant pressure.

Mass of the gas, m = (8 x 10³ cm³) x 0.00125 g · cm^-3 = 10 g.

Heat is gained by the gas at constant pressure.

Q = \(m c_p\left(t_f-t_i\right)\)

or, \(c_p=\frac{Q}{m\left(t_f-t_i\right)}=\frac{30}{10 \times(15-0)}=0.2 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

Now, mass of 8L 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, \(C_p=M c_p=28 \times 0.2 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

Molar-specific heat at constant volume. \(C_v=C_p-R=(28 \times 0.2-2) \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

∴ Specific heat of the gas at constant volume, \(c_y=\frac{C_y}{M}=\frac{28 \times 0.2-2}{28}=0.2-\frac{2}{28}\)

= \(0.1286 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

First Law Of Thermodynamics Closed System Numerical Problems Class 11

Example 6. The temperature of 20 g of oxygen gas is raised from 50°C to 100°C

1. At constant volume and
2. At constant pressure. Find out the amount of heat supplied and the rise in internal energy in each case. Given, R = 2 cal · mol^-1 · K^-1 for oxygen, c = 0.155 cal · g^-1 · °C^-1

Solution:

1. Heat supplied at constant volume,

Q_v = mc_v(t_f– t_i)

= 20 x 0.155 x (100 -50) = 155 cal

Work done, W = 0

So, rise in internal energy,

U_f – U_i = Q – W – 155 – 0 = 155 caJ

= 155 x 4.2 J = 651 J.

2. Oxygen may be taken as an ideal gas. Internal energy is a function of temperature only.

i.e., \(U_f-U_i=155 \mathrm{cal}=651 \mathrm{~J}\).

⇒ \(C_v=M c_v=0.155 \times 32=4.96 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

M = molecular mass of oxygen = 32

∴ \(C_p=C_v+R=4.96+2=6.96 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

∴ \(20 \mathrm{~g} \text { oxygen }=\frac{20}{32} \mathrm{~mol} \text { oxygen }\)

Heat supplied at constant pressure,

= \(\frac{20}{32} \times 6.96 \times(100-50)\)

= 217.5 cal.