Q#1
A Gas kept in a container of finite conductivity is suddenly compressed. The process
(a) must be very nearly adiabatic
(b) must be very nearly isothermal
(c) may be very nearly adiabatic
(d) may be very nearly isothermal.
Answer: (c), (d).
Explanation: Since it is not clear how much is the conductivity of the container and how sudden the gas is compressed there may be two conditions.
First, the thermal conductivity is very low and the compression is very fast. In this situation, there will be negligible heat transfer through the container. the process may be very nearly adiabatic. Option (c).
Second, the conductivity is very high and the sudden compression is not very fast. In this case, the temperature approximately remains constant. The process is very nearly isothermal. Option (d).
Q#2
Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process,
(a) Q = 0,
(b) W = 0,
(c) Q ≠ W,
(d) Q =W.
Answer: (d).
Explanation: From the first law of thermodynamics, dQ = dU + dW.
In an isothermal process, the temperature remains constant and so does the internal energy U. So dU = 0. Thus, dQ = dW. Hence for the question, Q = W. Option (d) is correct.
Q#3
Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process,
(a) Q = 0
(b) W = 0,
(c) Q = W,
(d) Q ≠ W.
Answer: (a), (d).
Explanation: For an adiabatic process, there is no heat transfer, hence Q = 0.
Thus, the first law of thermodynamics for this process is,
0 = U + W
W = -U. W is not zero.
Hence the options (b) and (c) are not true. The options (a) and (d) are true.
Q#4
Consider the processes A and B, shown in figure (27-Q3). Is it possible that
(a) both the processes are isothermal
(b) both the processes are adiabatic
(c) A is isothermal and B is adiabatic
(d) A is adiabatic and B is isothermal.
Answer: (c)
Explanation: Both the processes start from the same state, hence both of them cannot be either isothermal or adiabatic. In that case, both will traverse same path. So, the options (a) and (b) are not true.
Since one of them is isothermal and the other adiabatic, the steeper slope path B will be adiabatic and the other A will be isothermal. Hence the option (c) is true.
Q#5
Three identical adiabatic containers A, B and C contain helium, neon and oxygen respectively at equal pressure. The gasses are pushed to half their initial volumes.
(a) the final temperatures in the three containers will be the same.
(b) The final pressures in the three containers will be the same.
(c) The pressures of helium and neon will be the same but that of oxygen will e different.
(d) the temperatures of helium and neon will be the same but that of the oxygen will be different.
Answer: (c), (d).
Explanation: Helium and Neon are monoatomic gases for which Cₚ/Cᵥ = ɣ = 1.67, but oxygen is a diatomic gas for which ɣ = 1.4.
For an adiabatic process pVɣ = constant or TV(ɣ-1) =constant.
Hence the changes for helium and neon will be the same but for oxygen, it will be different. So the options (c) and (d) are true.
Q#6
A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increases by 1°C if 3.0 cal of heat is added to it. The gas may be
(a) helium
(b) argon
(c) oxygen
(d) carbon dioxide.
Answer: (a), (b).
Explanation: From the given situation at the constant volume one mole of the ideal gas requires 3.0 cal of heat to increase its temperature by 1°C which is by definition is the molar heat capacity of the gas at constant volume Cᵥ.
So, Cᵥ = 3.0 cal/mol-K = (3.0)(4.18 J/mol-K) = 12.5 J/mol-K.
We know that Cₚ/Cᵥ =ɣ,
Cₚ = ɣCᵥ
Also, Cₚ - Cᵥ = R
ɣCᵥ - Cᵥ = R
Cᵥ = R/(ɣ-1)
For a monoatomic gas ɣ = 1.67, thus
Cᵥ = R/0.67 = 1.5R = (1.5)(8.314 J/mol-K ) = 12.5 J/mol-K.
So, the given gas is monoatomic. Out of the four options only helium and argon are monoatomic because they are inert gases. Hence the options (a) and (b) are true.
Q#7
Four cylinders contain equal number moles of argon, hydrogen, nitrogen and carbon dioxide at the same temperature. The energy is minimum in
(a) argon
(b) hydrogen
(c) nitrogen
(d) carbon dioxide.
Answer: (a).
Explanation: Taking the energy to be zero at T = 0, the energy of gas at temperature T is given as
U = nCᵥT
Since in the given four cylinders, n and T are the same for all the gases, the energy will be minimum for the gas for which Cᵥ is minimum.
Since Cᵥ = R/(ɣ - 1), it will be minimum for which ɣ is maximum. We know that ɣ is maximum for a monoatomic gas and decreases as the atomicity increases. Out of the four options, only argon is monoatomic, and others are either diatomic or triatomic. So, Cᵥ is a minimum for argon. Thus, the option (a) is true.
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