Questions OBJECTIVE - I and Answer Specific Heat Capacities Of Gases HC Verma Part II
Q#1
Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If CA and CB be the molar heat capacities for the two processes,
(a) CA = CB
(b) CA < CB
(c) CA > CB
(d) CA and CB cannot be defined.
Answer: (c)
Explanation: From the statement of the first law of thermodynamics,
ΔQ = ΔU + ΔW.
Since the temperature rises through the same amount in both the processes, ΔU is the same for both the processes.
Given (ΔW)ₐ = 2(ΔW)ᵦ
Hence (ΔQ)ₐ > (ΔQ)ᵦ
Molar heat capacity C ={ΔQ/n.ΔT}
n.ΔT is the same in both the processes, hence, CA > CB. Option (c).
Q#2
For a solid with a small expansion coefficient,
(a) Cₚ - Cᵥ = R
(b) Cₚ = Cᵥ
(c) Cₚ is slightly greater than Cᵥ
(d) Cₚ is slightly less than Cᵥ.
Answer: (c)
Explanation: Since the expansion coefficient is small, there will be change in the volume but very small. So, a very small amount of heat supplied will be used for the work done in expansion. Hence Cₚ is slightly greater than Cᵥ.
Option (c).
Q#3
The value of Cₚ - Cᵥ is 1.00R for a gas sample in state A and 1.08R in state B. Let pA, pB denote the pressures and TA and TB denote the temperatures of the states A and B respectively. Most likely
(a) pA < pB and TA > TB
(b) pA > pB and TA < TB
(c) pA = pB and TA < TB
(d) pA > pB and TA = TB
Answer: (a)
Explanation: The relation Cₚ - Cᵥ =1.00R shows that it behaves as an ideal gas at state A. But at state B, Cₚ - Cᵥ =1.08R, so in this state, the gas is not behaving as an ideal gas but as real gas. A real gas behaves nearly as an ideal gas when it is at low pressure and high temperature. Hence in state A the given sample of the gas has low pressure and high temperature. So, the option (a) is true.
Q#4
Cᵥ and Cₚ denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively. Which of the following is a universal constant?
(a) Cₚ/Cᵥ
(b) CₚCᵥ
(c) Cₚ - Cᵥ
(d) Cₚ + Cᵥ.
Answer: (c)
Explanation: The product and sum of Cₚ and Cᵥ are not constant, so the options (b) and (d) are not true. Cₚ/Cᵥ=ɣ is a constant but not a universal constant because ɣ depends on the atomicity of the gas, so the option (a) is also not true. Cₚ - Cᵥ = R, which is the universal gas constant with a value of 8.314 J/mol-K. Hence the option (c) is true.
Q#5
70 calories of heat are required to raise the temperature of 2 mol of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is
(a) 30 calories
(b) 50 calories
(c) 70 calories
(d) 90 calories.
Answer: (b)
Explanation: ΔQ = nCₚΔT and
ΔQ' = nCᵥΔT.
Here ΔQ = 70 cal =70(4.2 J), n = 2 mol, ΔT = 5°K,
so, Cₚ = 70{4.2/(2 x 5) = 7(4.2) J/mol-K.
Now Cᵥ = Cₚ - R = (7)(4.2) - 8.3 = 29.4 - 8.3 = 21.1 J/mol-K.
Now ΔQ' = nCᵥΔT = 2(21.1)(5) = 211 J = 211/4.2 cal ≈ 50 cal. Hence option (b).
Note: you may not convert cal into J and write the unit of R as cal/mol-K.
Q#6
Figure (27-Q1) shows a process on a gas in which pressure and volume both change. The molar heat capacity for this process is C.
(a) C = 0
(b) C = Cᵥ
(c) C > Cᵥ
(d) C < Cᵥ
Answer: (c)
Explanation: Suppose the given process takes the state of the gas from A to B, the difference in temperature is dT and the heat involved is dQ. The difference in internal energy = dU.
dQ = nCdT
If the work done in this process = dW, then from the first law of thermodynamics, dQ = dU + dW ---- (i)
Now consider a process on the same amount of gas at a constant volume that takes the temperature of gas through the same difference. Hence the heat supplied in this process = dQ' = nCᵥdT
The work done in this process is zero due to constant volume. Hence dU = dQ'.
From (i), dQ = dU + dW
dQ = dQ' + dW
nCdT = nCᵥdT + dW
C = Cᵥ +dW/ndT
The area under p-V diagram in the given figure is positive hence C > Cᵥ. Option (c) is true.
Q#7
The molar heat capacity for the process shown in figure (27-Q2) is
(a) C = Cₚ
(b) C = Cᵥ
(c) C > Cᵥ
(d) C = 0.
Answer: (d)
Explanation: In the given figure
p = K/Vɣ
pVɣ = K which shows that it is an adiabatic process in which the heat supplied = zero. The molar heat capacity is defined as the amount of heat required per mol of the gas to raise the temperature through 1K. Since no heat is being given here to raise the temperature, the molar heat capacity is zero. Option (d).
Q#8
In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about
(a) 0.25%
(b) 0.5%
(c) 0.7%
(d) 1%
Answer: (b)
Explanation: In an isothermal process pV = constant. Hence if p increases by 0.5%, V will decrease by 0.5% to keep the product of the new values of p and V constant. Hence the option (b).
Q#9
In an adiabatic process on a gas with ɣ = 1.4, the pressure is increased by 0.5%. The volume decreases by about
(a) 0.36%
(b) 0.5%
(c) 0.7%
(d) 1%.
Answer: (a)
Explanation: In an adiabatic process, pVˠ = K. So if p is increased by 0.5%, Vˠ should be decreased by 0.5% to keep the product constant. Hence now, V'ˠ =Vˠ/1.005
V' = V/(1.005)(1/ˠ) =V/(1.005)(1/1.4) = V/1.0036
Which means the volume decreases by 0.36%.
The option (a) is true.
Q#10
Two samples A and B are initially kept in the same state. Sample A is expanded through an adiabatic process and sample B through an isothermal process. The final volumes of the samples are the same. The final pressures in A and B are pA and pB respectively.
(a) pA > pB
(b) pA = pB
(c) pA< pB
(d) The relation between pA and pB cannot be deduced.
Answer: (c)
EXPLANATION: In the adiabatic process pVˠ = K and in the isothermal process pV = K'.
Now pVˠ = pA V'ˠ and pV = pBV'
Dividing we get, V(γ-1) =(pA/pB)V'(γ-1)
Hence pA/pB = (V/V')(γ-1)
The gas is expanded, so V' > V, so
pA/pB < 1
pA < pB.
Option (c) is true.
Q#11
Let Tₐ and Tᵦ be the final temperatures of samples A and B respectively in the previous question.
(a) Tₐ < Tᵦ
(b) Tₐ = Tᵦ
(c) Tₐ > Tᵦ
(d) The relation between Tₐ and Tᵦ cannot be deduced.
Answer: (a)
Explanation: Since the sample-B goes through the isothermal process, its initial and final temperature will be the same and equal to Tᵦ. Initially, both the samples are at the same temperature hence the initial temperature of the sample-A = Tᵦ. This sample expands adiabatically hence the work done by the gas is at the expense of its internal energy. Hence the final temperature of this sample will decrease, so Tₐ < Tᵦ. Option (a) is true.
Q#12
Let ΔWₐ and ΔWᵦ be the work done by systems A and B respectively in the previous question.
(a) ΔWₐ > ΔWᵦ
(b) ΔWₐ = ΔWᵦ
(c) ΔWₐ < ΔWᵦ
(d) The relation between ΔWₐ and ΔWᵦ cannot be deduced.
Answer: (c)
Explanation: For the sample-A which goes through adiabatic process, ΔQ = 0. Since ΔQ = ΔU + ΔW, here,
0 = ΔU + ΔWₐ
ΔWₐ = -ΔU
For sample B, ΔQ = ΔU' + ΔWᵦ
Since the process on B is isothermal the temperature does not change.
Thus, there is no change in the internal energy. ΔU' = 0. So, ΔWᵦ = ΔQ.
ΔWₐ has a negative value while ΔWᵦ has a positive value. Clearly, ΔWₐ < ΔWᵦ. Hence the option (c) is true.
Q#13
The molar heat capacity of oxygen gas at STP is nearly 2.5R. As the temperature is increased, it gradually increases and approaches 3.5R. The most appropriate reason for this behavior is that at high temperatures
(a) oxygen does not behave as an ideal gas
(b) oxygen molecules dissociate in atoms
(c) the molecules collide more frequently
(d) molecular vibrations gradually become effective.
Answer: (d)
Explanation: As the temperature goes high the molecular vibrations gradually become effective. The molecules start vibrating about their mean positions.
So, a part of the heat given is utilized for this vibration and more heat is required to raise the temperature through the same degree than at the STP. Hence the molar heat capacity is more at high temperatures. The option (d) is true.
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