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Q. 44. Demonstrate that the process in which the work performed by an ideal gas is proportional to the corresponding increment of its internal energy is described by the equation pV^{n} = const, where n is a constant.
Solution. 44. According to the problem : A α U or dA = aU (where a is proportionality constant)
or, (1)
From ideal gas law, pV= v R T, on differentiating
pdV + Vdp = v RdT (2)
Thus from (1) and (2)
or,
Dividing both the sides by pV
On integrating n In V + In p = In C (where C is constant)
or,
Q. 45. Find the molar heat capacity of an ideal gas in a polytropic process pV^{n} = const if the adiabatic exponent of the gas is equal to γ. At what values of the polytropic constant n will the heat capacity of the gas be negative?
Solution. 45. In the polytropic process work done by the gas
(where T_{i} and T_{f} are initial and final temperature of the gas like in adiabatic process)
and
By the first law of thermodynamics Q = ΔU + A
According to definition of molar heat capacity when number of moles v = 1 and ΔT = 1 then Q = Molar heat capacity.
Here,
Q. 46. In a certain polytropic process the volume of argon was increased α = 4.0 times. Simultaneously, the pressure decreased β = 8.0 times. Find the molar heat capacity of argon in this process, assuming the gas to be ideal.
Solution. 46. Let the process be polytropic according to the law pV^{n} = constant
Thus,
So,
In the polytropic process molar heat capacity is given by
So,
Q. 47. One mole of argon is expanded polytropically, the polytropic constant being n = 1.50. In the process, the gas temperature changes by ΔT = — 26 K. Find:
(a) the amount of heat obtained by the gas;
(b) the work performed by the gas
Solution. 47. (a) Increment of internal energy for ΔT, becomes
From first law of thermodynamics
(b) Sought work done,
Q. 48. An ideal gas whose adiabatic exponent equals y is expanded according to the law p = αV , where a is a constant. The initial volume of the gas is equal to V_{0}. As a result of expansion the volume increases η times. Find:
(a) the increment of the internal energy of the gas;
(b) the work performed by the gas;
(c) the molar heat capacity of the gas in the process.
Solution. 48. LaW 0f the process is p = α V or pV^{1} = α
so the process is polytropic of index n =  1
As p = αV so, P_{i}  αV_{0 }and p_{f} = α η V_{0}
(a) Increment of the internal energy is given by
(b) Work done by the gas is given by
(c) Molar heat capacity is given by
Q. 49. An ideal gas whose adiabatic exponent equals γ is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Find:
(a) the molar heat capacity of the gas in this process;
(b) the equation of the process in the variables T, V;
(c) the work performed by one mole of the gas when its volume increases η times if the initial temperature of the gas is T_{0}.
Solution. 49.
where C_{n} is the molar heat capacity in the process. It is given that
So,
(b) By the first law of thermodynamics, dQ  dU + dA,
or,
So,
or,
(c)
But from part (a), we have
Thus
From part (b); we know
So, (where T is the final temperature)
Work done by the gas for one mole is given by
Q. 50. One mole of an ideal gas whose adiabatic exponent equals y undergoes a process in which the gas pressure relates to the temperature as p = aT^{α}, where a and α are constants. Find:
(a) the work performed by the gas if its temperature gets an increment ΔT;
(b) the molar heat capacity of the gas in this process; at what value of α will the heat capacity be negative?
Solution. 50. Given p = a T^{α} (for one mole of gas)
So,
or,
Here polytropic exponent
(a) In the poly tropic process for one mole of gas :
(b) Molar heat capacity is given by
Q. 51. An ideal gas with the adiabatic exponent γ undergoes a process in which its internal energy relates to the volume as U = aV^{α}, where a and α are constants. Find:
(a) the work performed by the gas and the amount of heat to be transferred to this gas to increase its internal energy by ΔU;
(b) the molar heat capacity of the gas in this process.
Solution. 51.
or,
or,
or
So polytropric index n = 1  α
(a) Work done by the gas is given by
Hence
By the first law of thermodynamics, Q = ΔU+A
(b) Molar heat capacity is given by
Q. 52. An ideal gas has a molar heat capacity C_{v} at constant volume. Find the molar heat capacity of this gas as a function of its volume V, if the gas undergoes the following process:
Solution. 52. By the first law .of thermodynamics
Molar specific heat according to definition
We have
After differentiating, we get
So,
Hence
(b)
So,
Q. 53. One mole of an ideal gas whose adiabatic exponent equals γ undergoes a process p = p_{0} + α /V, w here P_{0} and α are positive constants. Find:
(a) heat capacity of the gas as a function of its volume;
(b) the internal energy increment of the gas, the work performed by it, and the amount of heat transferred to the gas, if its volume increased from V_{1} to V_{2}.
Solution. 53. Using 2.52
(for one mole of gas)
Therefore
Hence
(b) Work done is given by
By the first law of thermodynamics Q = ΔU +A
Q. 54. One mole of an ideal gas with heat capacity at constant pressure C_{p} undergoes the process T = T_{0} + αV, where T_{0} and α are constants. Find:
(a) heat capacity of the gas as a function of its volume;
(b) the amount of heat transferred to the gas, if its volume increased from V_{1} to V_{2}.
Solution. 54. (a) Heat capacity is given by
(see solution of 2.52)
We have
After differentiating, we get,
Hence
(b)
for one mole of gas
Now
By the first law of thermodynamics Q = ΔU + A
Q. 55. For the case of an ideal gas find the equation of the process (in the variables T, V) in which the molar heat capacity varies as:
(a) C = C_{v} + αT; (b) C = C_{v} + βV; (c) C = C_{v} + ap, where α, β, and a are constants.
Solution. 55. Heat capacity is given by
Integrating both sides, we get is a constant.
Or,
and
or,
Integrating both sides, we get
So,
So,
or
or,
or V  a T = constant
Q. 56. An ideal gas has an adiabatic exponent γ. In some process its molar heat capacity varies as C = α/T, where α is a constant. Find:
(a) the work performed by one mole of the gas during its heating from the temperature T_{0} to the temperature η times higher;
(b) the equation of the process in the variables p, V.
Solution. 56. (a) By the first law of thermodynamics A = Q  ΔU
or,
Given
So,
(b)
Given
or,
or,
or,
Integrating both sides, we get
or,
or,
or,
or,
Q. 57. Find the work performed by one mole of a Van der Waals gas during its isothermal expansion from the volume V_{1} to V_{2} at a temperature T.
Solution. 57. The work done is
Q. 58. One mole of oxygen is expanded from a volume V_{1} = 1.00 1 to V_{2} = 5.0 l at a constant temperature T = 280 K. Calculate:
(a) the increment of the internal energy of the gas:
(b) the amount of the absorbed heat.
The gas is assumed to be a Van der Waals gas.
Solution. 58. (a) The increment in the internal energy is
But from second law
On the other hand
or,
So,
(b) From the first law
Q. 59. For a Van der Waals gas find:
(a) the equation of the adiabatic curve in the variables T, V;
(b) the difference of the molar heat capacities C_{p}, — C_{v} as a function of T and V.
Solution. 59. (a) From the first law for an adiabatic
dQ = dU + pd V = 0
From the previous problem
So,
This equation can be integrated if we assume that Cv and b are constant then
or,
(b) We use
Now,
So along constant p,
Thus
On differentiating,
or,
and
Q. 60. Two thermally insulated vessels are interconnected by a tube equipped with a valve. One vessel of volume V_{1} = 10 l contains v = 2.5 moles of carbon dioxide. The other vessel of volume V_{2} = 100 l is evacuated. The valve having been opened, the gas adiabatically expanded. Assuming the gas to obey the Van der Waals equation, find its temperature change accompanying the expansion.
Solution. 60. From the first law
as the vessels are themally insulated.
As this is free expansion,
But
So,
or,
Substitution gives ΔT =  3 K
Q. 61. What amount of heat has to be transferred to v = 3.0 moles of carbon dioxide to keep its temperature constant while it expands into vacuum from the volume V_{1} = 5.0 l to V_{2} = 10 l ? The gas is assumed to be a Van der Waals gas.
Solution. 61. (as A = 0 in free expansion).
So at constant temperature.
= 0.33 kJ from the given data.
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