HC Verma Questions and Solutions: Chapter 27: Specific Heat Capacities of Gases- 1 | HC Verma Solutions - JEE PDF Download

According to the first law of thermodynamics, ΔQ = ΔU + ΔW, where Δ Q  is the heat supplied to the system when ΔW work is done on the system and ΔU  is the change in internal energy produced. Since the temperature rises by the same amount in both processes, change in internal energies are same, i.e. ΔU_A = ΔU_B.
But as , ΔW_A =ΔW_B this gives ΔQ_A = 2ΔQ_B.
Now, molar heat capacity of a gas, C =  where Δ Q/n is the heat  supplied to a mole of gas and ΔT is the change in temperature produced. As ΔQ_A = 2ΔQ_B, C_A > C_B. 

For a solid with a small expansion coefficient, work done in a process will also be small. Thus, the specific heat depends slightly on the process. Therefore, C_p is slightly greater than C_v.

C_p − C_v  = R for the gas in state A, which means it is acting as an ideal gas in that state, whereas C_p − C_v = 1.08R in state B, i.e. the behaviour of the gas is that of a real gas in that state. To be an ideal gas, a real gas at STP should be at a very high temperature and low pressure. Therefore, P_A < P_B and T_A > T_B where P_A and P_B denotes the pressure and T_A and T_B denotes the temperature of system A and B reepectively.

For an ideal gas, C_p − C_v = R , where C_v and C_p denote the molar heat capacities of an ideal gas at constant volume and constant pressure, respectively  and R is the gas constant whos value is 8.314 J/K. Therefore, C_p − C_v is a constant. On the other hand, the ratio of these two varies as the atomicity of the gas changes. Also, their sum and product are not constant.

It is given that 70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. Also, specific heat at constant pressure,

For an ideal gas,
C_P - C_V = R = 8.314 J - mol ^-1 K^-1 ≃ 2 calories mol^-1K^-1 
⇒ C_V = C_P -R
⇒ C_V = (7-2) calories - mol^-1 K^-1
⇒ C_V = 5 calories - mol^-1 K^-1

⇒ Δ Q = 5 × 2 × (35-30)
⇒ Δ Q = 5 × 2 × 5
⇒ Δ Q = 50 calories
Therefore, 50 calories need to be supplied to raise the temperature of 2 moles of gas from 30-35ºC at constant volume.

C > C​v
Consider two processes AB and ACB; let W be the work done. C is the molar heat capacity of process AB. Process ACB can be considered as the sum of the two processes, AC and CB. The molar heat capacity of process AC is C_p, as pressure is constant in this process and the molar heat capacity of process CB is C_v, as volume is constant in it. 
Internal energy, U, is a state function, i.e. it doesn't depend on the path followed. Therefore,
U_AB = U_ACB
W_AB > W_ACB
Work done in the p-V diagram is the area enclosed under the curve.
⇒ W_AB + U_AB > W_ACB + U_ACB
⇒ C > C_V + C_P
Molar heat capacity is the heat supplied per mole to change the temperature by a degree Kelvin and according to the first law of thermodynamics, dQ = dU + dW, where dQ is the heat supplied to the system in a process.
⇒ C > C_V 

C = 0.he defined process is 

such that the process is adiabatic in which there's no heat supplied to the system, i.e. Q= 0. Molar heat capacity is the amount of heat supplied to the system per mole to produce a degree change in temperature. Also, in an adiabatic process, no heat exchange is allowed. So, molar heat capacity equals zero, i.e. C = 0.

Let p and p' be the initial and final pressures of the system and V and V' be the initial and final volumes of the system.  p' is 0.5% more than p and the process is isothernal. So, pV = k = p'V' =  constant. Therefore, 
p^V = p^'V'

= -0.49 %
So, volume V' decreases by about 0.5% of V.

Let p and p, be the initial and final pressures of the system and V and V, be the initial and final volumes of the system. p, is 0.5% more than p and the process is adiabatic. So,

⇒ V' = 0.99644 V
⇒ V' =V -0.00356 V
Therefore, V' is 0.36 % less than V.

Let the initial states of samples A and B be i and the final states of samples B and A be fand f', respectively. Let the final volumes of both be V_o. As sample A is expanded through an adiabatic process, its curve in the p-V diagram is steeper than that of sample B, which is expanded through an isothermal process. Therefore, from the p-V diagram, p_A < p_B.

As sample B is undergoing expansion through an isothermal process, its initial and final temperatures will be same, i.e. T_b. On the other hand, sample A is at the same initial state as B, such that the initial temperature of A is ​T_b and it is expanding through an adiabatic process in which no heat is supplied. Therefore, sample A will expand at the cost of its internal energy and its final temperature will be less than its initial temperature.
This implies that  T_a < T_b.

In the p-V diagram, the area under the curve w.r.t the V axis is equal to the work done by the system. Since the area under the isotherm is greater than that under the adiabat, the work done by system A is less than that done by system B. Hence,  ∆W_a < ∆W_b.

Molar specific heat capacity has direct dependence on the degree of freedom of gas molecules. As temperature is increased, the gas molecules start vibrating about their mean position, leading to change (increase) in the degree of freedom and, hence, increasing molar heat capacity.

Due to sudden compression, the gas did not get sufficient time for heat exchange. So, no heat exchange occurred. Therefore, the process may be adiabatic. For any process to be isothermal, its temperature should remain constant, i.e. pressure and volume should change simultaneously while their product (temperature) should be constant.

In an isothermal process, temperature of the system stays constant, i.e. there's no change in internal energy. Thus, U = 0, where U denotes the change in internal energy of the system. According to the first law of thermodynamics, heat supplied to the system is equal to the sum of change in internal energy and work done by the system, such that Q = U + W. As U = 0, Q = W.

In an adiabatic process, no heat is supplied to the system; so, Q = 0. According to the first law of thermodynamics, heat given to any system is equal to the sum of the change in internal energy and the work done on the system. So, Q = W+U and as Q = 0, W = -U and  Q ≠ W.

The slope of an adiabatic process is greater than that of an isothermal process. Since Aand B are initiated from the same initial state, both cannot be isothermal or adiabatic, as they would be overlapping. But the curve of process B is steeper than the curve of process A. Hence, A is isothermal and B is adiabatic.

Adiabatic process is expressed as 
p^Vγ = constant or
T^Vγ-1 = constant ,

is the ratio of molar heat capacities at constant pressure and volume. 
We know that γ is equal to 1.67 and 1.40 for a monatomic gas and a diatomic gas, respectively. Helium and neon are monatomic gases and oxygen is a diatomic gas. Therefore, changing the state of the gases, i.e. reducing the volume will lead to identical changes in temperature and pressure for helium and neon and that will be different for oxygen. 

The temperature of one mole of a gas kept in a container of fixed volume is increased by 1 degree Celsius if 3 calories, i.e. 12.54 J of heat is added to it. So, its molar heat capacity, C​_v = 12.54 J   JK^-1 mol^-1, as molar heat capacity at fixed volume is the heat supplied to a mole of gas to increase its temperature by a degree. For a monatomic gas,  Among the given gases, only helium and argon are inert and, hence, monoatomic. Therefore, the gas may be helium or argon.

The energy of a gas is measured as C_vT. All the four cylinders are at the same temperature but the gases in them have different values of  C_v, such that it is least for the monatomic gas and keeps on increasing as we go from monatomic to tri-atomic. Among the above gases, argon is monatomic, hydrogen and nitrogen are diatomic and carbon dioxide is tri-atomic. Therefore, the energy is minimum in argon.