Solution:

Molar specific heat capacity is the heat energy required to raise the temperature of 1 mole of a substance by `1 K` and expressed in `J moi^(-1) K^(-1.)`

`C= (Q)/(1mole1K)`

Depending on the condition that whether volume or pressure is constant, molar specific heat is written as `Cv` and `CP'` Relation between `C_p` and `C_p` Suppose one mole of a gas is heated so that its temperature rises by `dT.` Heat supplied `= 1 x Cv x dT = CvdT ... (i)` Since the volume is constant, the gas will not perform external work in accordance with the first law of thermodynamics and the heat supplied will be just equal to the increase in the internal energy of the gas.

`:. dU = C_vdT.....(ii)`

Let the gas be heated at constant pressurento again increase its temperature by `dT,` and `dQ` be the amount of heat supplied, therefor
`dQ = 1xxC_p xxdT=C_pdT.(iii)`

The heat supplied at a constant pressure increases the temperature by `dT` hence increases its internal energy by `dU = CpdT` as well as enables the gas to perform work `dW.`

`dW= PdV`

From the' first law of thermodynamics, we have

`dQ = dU + dW`

Substituting the values, we get,

`C_PdT = C_pdT + PdV`

But `PV = RT` (For one mole of the gas)

or `PdV = RdT`

`. . C_PdT = C_vdT + RdT`

or `C_p-C_v=R`

This is the relation between two principal specific heats of the gas when `C_P, C_v` and R are measured in the units of either heat or of work.

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Solution:

`"Kelvin-Planck Statement of Second Law of Thermodynamics."` It is impossible to realise a heat engine which works in a cyclic process performing the single job of taking heat from a body at a constant temperature and converting it completely into mechanical work. The law stresses the fact that work done on the system can facilitate to convert heat into work however with some unavoidable loss. `"Clausius Statement :"` It is impossible for a self~acting machine, unaided by any external agency, to transfer heat from a body at lower temperature to another at higher temperature.

Efficiency in forward cycle `= text(Work done)/("Heat supplied")`

`(Q_1-Q_2)/(Q_1) = 1(Q_2)/(Q_1) = 1(T_2)/(T_1)`

Coefficient of performance `= text(Heat drawn from sink)/text(Work on the system)`

`(Q_1)/(Q_1-Q_2) = (T_2)/(T_2)/(T_1)`

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Solution:

First Law of Thermodynamics. It is based on the conservation of energy. The total heat energy change in any system is the sum of the internal energy change and the work done, i.e., `dQ = dU + Jw` where `dU ->` internal energy change and `dW = PdV` is the work done by/on the system, dU is a state function and depends on `dT.` `(dU = nCvdT). dW` depends on the path followed and so, is different in various processes. Relationship between `CP` and `Cv`. Suppose one mole of a gas is heated at constant volume so that its temperature rises by `dT.` Heat supplied `= 1 xx Cv xx dT = CvdT ... (i)` Since the volume is constant, the gas will not perform external work in accordance with the first law of thermodynamics and the heat supplied will be just equal to the increase in the internal energy of the gas. `. . dU = C_vdT ... (ii)` Let the gas be heated at constant pressure to again increase its temperature by `dT,` and `dQ` be the amount of heat supplied, therefore, `dQ = 1 xx C_p xx dT = C_pdT` ... (iiii)
The heat supplied at a constant pressure increases the temperature by dT hence increases its internal energy by `dU = C_udT` as well as enables the gas to perform work `dW.` `dW = PdV ... (iv)` From the first law of thermodynamics, we have `dQ = dU + dW` Substituting the values, we get `C_PdT = C_udT + PdV` But `PV = RT` (For one mole of an ideal gas) or `PdV = RdT, :. C_PdT = C_vdT + RdT` or `C_p-C_v=R ... (v)` This is the relationship between two principal specific heats of the gas when `C_P, C_v` and R are measured in the units of either heat or of work.

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Solution:

Define molar specific heat. Write its units.Molar specific heat is defined as the amount of heat required to raise the temperature of one mole of a gas through 1 K at constant volume or at constant pressure. If the volume is constant, it is called the molar specific heat at constant volume. Similarly, if the pressure is constant, it is called the

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Solution:

(a) Adiabatic process : When no exchange of heat energy is possible between system and surrounding, the process is known as adiabatic. Work done in adiabatic process: Consider 1 mole of gas contained in a perfectly non-frictionless conducting cylinder fitted with a piston (non conducting). Let

P = PresE;ure of gas
V = Volume of gas
T = Temperature of gas

Compress the gas adiabatically by moving the piston inwards through a distance d`x.` Force acting on piston `= P xx A` Work done,

`dW = "Force xx distance"`
`= PA dx = PdV`

where `dV` is the change in volume of the gas. During the adiabatic expansion, total work done by the gas

`W = int_(V_1)^(V_2) PdV`

Adiabatic change is represented by `Pv^Y= K,` (a constant)

`P = K/V^gamma`

Hence, `W = int_(v_1)^(v_2) PdV`

`= K/(1 - gamma[V_(2)^(1) - V_(1)^(1-gamma)]`

`= 1/(1-gamma) [P_2V_(2)^(1_gamma)-P_2V_1^(gamma)V_(1)^(1-gamma)]`

`(P_1V_(1)^(gamma)- P_2V_(2)^(gamma)=K)`

`= 1/(1-gamma) [RT_2 - RT_1]`

(from ideal gas equation)

or `W = (R)/(1-gamma[T_2-T_1]`

This is the work done during adiabatic process.

(b) Given : `Q_1 = 6 xX 105 cal, T_1 = 500 K, T_2 = 400 K`

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Solution:

Temperature is a scalar physical quantity and is a property of all thermodynamic systems (in equilibrium states) such that temperature equality is a necessary and sufficient condition for thermal equilibrium.

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Solution:

Zeroth law defines temperature and first law defines internal energy.

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Solution:

`C_P:` The amount of energy required for 1 mole of a gas to raise its temperature by 1 K at constant pressure condition.

`C_V` :The amount of energy required for 1 mole of a gas to raise its temperature by 1 K at constant volume conditions.

`C_P > C_V`Since for constant pressure process, both volume and temperature are altered and for constant volume process, only temperature varies.

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Solution:

According to the first law of thermodynamics, the total heat energy change `dQ` is the sum of the internal energy change `dU` and work done `dW, i.e., dQ = dU + dW.` Heat energy given to the system is `+ve,` taken out is `-ve`. Internal energy change is `+ve` with increase in temperature. Work done is `+ve` if volume increases and `-ve` if volume decreases.

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Solution:

`"Kelvin-Planck Statement of Second Law of Thermodynamics."` It is impossible to realise a heat engine which works in a cyclic process performing the single job of taking heat from a body at a constant temperature and converting it completely into mechanical work. The law stresses the fact that work done on the system can facilitate to convert heat into work however with some unavoidable loss. `"Clausius Statement :"` It is impossible for a self~acting machine, unaided by any external agency, to transfer heat from a body at lower temperature to another at higher temperature.

Efficiency in forward cycle `= text(Work done)/("Heat supplied")`

`(Q_1-Q_2)/(Q_1) = 1(Q_2)/(Q_1) = 1(T_2)/(T_1)`

Coefficient of performance `= text(Heat drawn from sink)/text(Work on the system)`

`(Q_1)/(Q_1-Q_2) = (T_2)/(T_2)/(T_1)`

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Solution:

No engine working between two temperatures can be more efficient than a Carnot's reversible engine working between the same temperatures.

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Solution:

Law of equi-partition of energy states that every degree of freedom will provide the same amount of energy to the internal energy of the system, i.e , 1/2 RT Internal energy with mono, di and tri-atomic gas is,

`3/2R, 5R` and `7/2 R` respectively

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Solution:

According to the Dulong and Petit's law, the specific heat per mole of a chemically pure crystalline solid is approximately `6 cal moi^(-1) K^(-1)` or roughly `25 J mol e^(-1) K^(-1).` This law is obeyed with quite a good approximation for many substances at the room temperature.

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Solution:

According to the equipartition of energy, each degree of freedom will contribute an equal energy of `1/2` RT per mole In mono, di and triatomic gases hawing 3, 5 and 6 or 7 degrees offreedom, the internal energy will be `3/2 RT, 5/2` and `6/2 RT` or `7/2 RT.` Using `dU = nC_vdT` for mole, we get `C_v = 3/2 R, 5/2R` and `3R` or `7/2R` for the three gases respectively

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Solution:

It is a device that helps in maintaining a constant temperature. It consists of a bi-metallic strip which comprises of two thin strips of different materials (such as brass and invar) welded together along their lengths. On heating, this combination bends into an arc. It happens because brass has a higher coefficient of expansion than invar.

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Solution:

(i) Heat engine - One cannot devise a process whose only purpose is to convert heat into work.
(ii) Refrigerator- One cannot make heat to flow from a cold to a hot body by itself, without using any external agency.

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Solution:

According to first law of thermodynamics when some quantity of heat (dQ) is supplied to a system capable of doing external work, then
quantity of heat absorbed by the system (dQ) is equal to sum of increase in internal energy of system (dU) due to rise in temperature and external work done by system (dW) in expansion i.e.

`dQ = dU +dW`

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Solution:

For isothermal process, `PV =` constant
Differentiating, `VdP + PdV = 0`

`:. (dP)/(dV) = -P/V`

For adiabatic process, Pv = costant Dioffernting,
`V'dp + gammaPV^(gamma-1) dV = 0`

`(dP)/(dV) = (gammaP)/(V)`

Comparing the two ratios, we can say, slope of adiabatic process is `gamma` times the slope of isothermal process.

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Solution:

(i) We know `PV^(gamma) =` constant for adiabatic process.

Also, `PV = nRT. :. P = (nRT)/(V)`

Replacing P, we have `(nRT)/(V) V^(gamma) =` constant (or) `Tv^(gamma-1) =` constant

(ii) From `PV = nRT` , we have `V =(nRT)/(P)`

Replacing V, we have `V =(nRT)/(P)^(gamma) =` constant
`T^gamma) P^(1-gamma) =` constant

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