Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics
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81.
A Carnot engine is working between $${127^ \circ }C$$ and $${27^ \circ }C.$$ The increase in efficiency will be maximum when the temperature of
A
the source is increased by $${50^ \circ }C$$
B
the sink is decreased by $${50^ \circ }C$$
C
source is increased by $${25^ \circ }C$$ and that of sink is decreased by $${25^ \circ }C$$
D
both source and sink are decreased by $${25^ \circ }C$$ each.
Answer :
the sink is decreased by $${50^ \circ }C$$
$$\frac{{{T_1} - {T_2}}}{{\;{T_1}}}$$ is maximum in case (B).
82.
110 joules of heat is added to a gaseous system, whose internal energy is $$40\,J;$$ then the amount of external work done is
A
$$150\,J$$
B
$$70\,J$$
C
$$110\,J$$
D
$$40\,J$$
Answer :
$$70\,J$$
$$\eqalign{
& \Delta Q = \Delta U + \Delta W \cr
& \Rightarrow \Delta W = \Delta Q - \Delta U = 110 - 40 = 70\,J \cr} $$
83.
A gas is taken through the cycle $$A \to B \to C \to A,$$ as shown. What is the net work done by the gas?
A
$$2000\,J$$
B
$$1000\,J$$
C
Zero
D
$$-2000\,J$$
Answer :
$$1000\,J$$
According to question,
Net work done = Area enclosed in $$pV$$ curve i.e. $$\Delta ABC$$
Area of $$\Delta ABC = \frac{1}{2} \times 5 \times {10^{ - 3}} \times 4 \times {10^5}\,J$$
$$ = {10^3}J = 1000\,J$$
84.
A closed system undergoes a process $$1 \to 2$$ for which the values $${W_{1 - 2}}$$ and $${Q_{1 - 2}}$$ are $$50\,kJ$$ and $$- 20\,kJ$$ respectively. If the system is returned to state 1 and $${Q_{2 \to 1}}$$ is $$+ 10\,kJ$$ the work done $${W_{2 \to 1}}$$ is
A
$$40\,kJ$$
B
$$50\,kJ$$
C
$$-60\,kJ$$
D
$$-50\,kJ$$
Answer :
$$-60\,kJ$$
$$\eqalign{
& - 20 = \Delta U + 50 \Rightarrow \Delta U = - 70. \cr
& {\text{From}}\,\,2 \to 1,\,\Delta U = 70\,kJ. \cr
& {\text{Now}}\,\,10 = 70 + W \Rightarrow W = - 60\,kJ. \cr} $$
85.
A thermodynamic system undergoes cyclic process $$ABCDA$$ as shown in figure. The work done by the system in the cycle is
A
$${p_0}{V_0}$$
B
$$2{p_0}{V_0}$$
C
$$\frac{{{p_0}{V_0}}}{2}$$
D
zero
Answer :
zero
Work done in the cyclic process = Area bounded by the closed configuration = Area of closed configuration
$$ = \frac{1}{2} \times {V_0} \times {p_0} - \frac{1}{2} \times {V_0} \times {p_0} = 0\left( {{\text{zero}}} \right).$$
86.
One mole of a diatomic ideal gas undergoes a cyclic process $$ABC$$ as shown in figure. The process $$BC$$ is adiabatic. The temperatures at $$A, B$$ and $$C$$ are 400 $$K,$$ 800 $$K$$ and 600 $$K$$ respectively. Choose the correct statement :
A
The change in internal energy in whole cyclic process is
250 $$R.$$
B
The change in internal energy in the process $$CA$$ is 700 $$R.$$
C
The change in internal energy in the process $$AB$$ is $$- 350\,R.$$
D
The change in internal energy in the process $$BC$$ is $$- 500\,R.$$
Answer :
The change in internal energy in the process $$BC$$ is $$- 500\,R.$$
In cyclic process, change in total internal energy is zero.
$$\eqalign{
& \Delta {U_{{\text{cyclic}}}} = 0 \cr
& \Delta {U_{BC}} = n{C_v}\,\Delta T \cr
& = 1 \times \frac{{5\,R}}{2}\,\Delta T \cr} $$
Where, $${C_v}$$ = molar specific heat at constant volume.
For $$BC,\,\Delta T = - 200\,K$$
$$\therefore \,\,\Delta {U_{BC}} = - 500\,R$$
87.
A gas is compressed from a volume of $$2{m^3}$$ to a volume of $$1{m^3}$$ at a constant pressure of $$100\,N/{m^2}.$$ Then it is heated at constant volume by supplying $$150\,J$$ of energy. As a result, the internal energy of the gas:
A
increases by $$250\,J$$
B
decreases by $$250\,J$$
C
increases by $$50\,J$$
D
decreases by $$50\,J$$
Answer :
increases by $$250\,J$$
As we know,
$$\eqalign{
& \Delta Q = \Delta U + \Delta W\,\,\left( {{{\text{1}}^{{\text{st}}}}{\text{ law of thermodynamics}}} \right) \cr
& \Rightarrow \Delta Q = \Delta U + P\Delta V \cr
& {\text{or}}\,\,150 = \Delta U + 100\left( {1 - 2} \right) = \Delta U - 100 \cr
& \therefore \Delta U = 150 + 100 = 250\,J \cr} $$
Thus the internal energy of the gas increases by $$250\,J$$
88.
In thermodynamic processes which of the following statements is not true ?
A
In an adiabatic process the system is insulated from the surroundings
B
In an isochoric process pressure remains constant
C
In an isothermal process the temperature remains constant
D
In an adiabatic process $$p{V^\gamma } = {\text{constant}}$$
Answer :
In an isochoric process pressure remains constant
For an adiabatic process, there should not be any exchange of heat between the system and its surroundings. All walls of the container must be perfectly insulated. In adiabatic changes, gases obey Poisson's law, i.e., $$p{V^\gamma } = {\text{constant}}.$$ In an isochoric process, volume remains constant and for isobaric process, pressure remains constant.
89.
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure $${P_i} = {10^5}$$ $$Pa$$ and volume $${V_i} = {10^{ - 3}}{m^3}$$ changes to a final state at $${P_f} = \left( {\frac{1}{{32}}} \right) \times {10^5}$$ $$Pa$$ and $${V_f} = 8 \times {10^{ - 3}}{m^3}$$ in an adiabatic quasi - static process, such that $${P^3}{V^5}$$ = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at $${P_i}$$ followed by an isochoric (isovolumetric) process at volume $${V_f}.$$ The amount of heat supplied to the system in the two - step process is approximately
A
$$112\, J$$
B
$$294\, J$$
C
$$588\, J$$
D
$$813\, J$$
Answer :
$$588\, J$$
$$\eqalign{
& {P^3}{V^5} = {\text{constant}} \cr
& \Rightarrow \,\,P{V^{\frac{5}{3}}} = {\text{constant}} \cr
& \Rightarrow \,\,\gamma = \frac{5}{3} \cr} $$
⇒ monoatomic gas
For adiabatic process
$$\eqalign{
& W = \frac{{{P_f}{V_f} - {P_i}{V_i}}}{{1 - \gamma }} \cr
& = \frac{{\frac{1}{{32}} \times {{10}^5} \times 8 \times {{10}^{ - 3}} - {{10}^5} \times {{10}^{ - 3}}}}{{1 - \frac{5}{3}}} \cr
& \therefore \,\,W = \frac{{25 - 100}}{{\frac{{\left( {3 - 5} \right)}}{3}}} \cr
& = \frac{{75 \times 3}}{2} \cr
& = 112.5\,J \cr} $$
From first law of thermodynamics $$q = \Delta U + w$$
$$\eqalign{
& \therefore \,\,\Delta U = - w \cr
& \therefore \,\,\Delta U = - 112.5\,J \cr} $$
Now applying first law of thermodynamics for process
1 & 2 and adding $${q_1} + {q_2} = \Delta U + {P_i}\left( {{V_f} - {V_i}} \right)$$
$$\eqalign{
& = - 112.5 + {10^5}\left( {8 - 1} \right) \times {10^{ - 3}} \cr
& = 587.55 \cr} $$
90.
The specific heat of a gas in an isothermal process is
A
infinite
B
zero
C
negative
D
remains constant
Answer :
infinite
Inisothermal process temperature remains constant.
i.e., $$\Delta T = 0.$$ Hence according to
$$C = \frac{Q}{{m\Delta T}} \Rightarrow {C_{_{{\text{iso}}}}} = \infty $$