Practice Start Test

Heat and Thermodynamics

NEET Physics · 97 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
Which of the following relations does not give the equation of an adiabatic process, where terms have their usual meaning?
A P 1−-γ\gamma T γ\gamma = constant
B PV γ\gamma = constant
C TV γ\gamma −-1 = constant
D P γ\gamma T 1−-γ\gamma = constant
Correct Answer
Option D
Solution

For an adiabatic process,

PVγP{V^\gamma }

= constant …(i) According to ideal gas equation

PV=nRT⇒P=nRTV{\rm{PV = nRT }} \Rightarrow P = {{nRT} \over V}

Putting value of P in (i), we get

nRTVVγ{{nRT} \over V}{V^\gamma }

= constant; ∴\therefore

TVγ−1T{V^{\gamma - 1}}

= constant Again from the ideal gas equation

V=nRTPV = {{nRT} \over P}

Putting value of V in (i), we get

P(nRTP)γP{\left( {{{nRT} \over P}} \right)^\gamma }

= constant;

P1−γTγ{P^{1 - \gamma }}{T^\gamma }

= constant

Q52
Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature T 1 and rejects the heat to the sink at temperature T. The second engine B receives the heat at temperature T and rejects to its sink at temperature T 2 . For what values of T the efficiencies of the two engines are equal
A T1−T22{{{T_1} - {T_2}} \over 2}
B T1T2{T_1}{T_2}
C T1T2\sqrt {{T_1}{T_2}}
D T1+T22{{{T_1} + {T_2}} \over 2}
Correct Answer
Option C
Solution

Efficiency of engine A,

η1=1−TT1{\eta _1} = 1 - {T \over {{T_1}}}

Efficiency of engine B,

η2=1−T2T{\eta _2} = 1 - {{{T_2}} \over T}

Here,

η1=η2{\eta _1} = {\eta _2}
∴TT1=T2T⇒T=T1T2\therefore {T \over {{T_1}}} = {{{T_2}} \over T} \Rightarrow T = \sqrt {{T_1}{T_2}}
Q53
A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is Ua=10 J{U_a} = 10\,J. Along the path adc the amount of heat absorbed dQ 1 == 50 J and the work obtained dW 1 == 20 J whereas along the path abc the heat absorbed dQ 2 = 36 J. The amount of work allong the path abc is
A 10 J
B 12 J
C 36 J
D 6 J
Correct Answer
Option D
Solution

From first law of thermodynamics

Qadc=ΔUadc+Wadc{Q_{adc}} = \Delta {U_{adc}} + {W_{adc}}
50 J=ΔUadc+20J50\,J = \Delta {U_{adc}} + 20J
ΔUadc=30J\Delta {U_{adc}} = 30J

Again

Qabc=ΔUabc+Wabc{Q_{abc}} = \Delta {U_{abc}} + {W_{abc}}
Wabc=Qabc−ΔUabc{W_{abc}} = {Q_{abc}} - \Delta {U_{abc}}
Wabc=36 J−30 J{W_{abc}} = 36\,J - 30\,J
Wabc=6 J{W_{abc}} = 6\,J
Q54
In a vessel, the gas is at pressure P. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be
A 2P
B P
C P/2
D 4P
Correct Answer
Option A
Solution

As

P=13mNVvrms2P = {1 \over 3}{{mN} \over V}v_{rms}^2

...(i) where m is the mass of each molecule, N is the total number of molecules, V is the volume of the gas.

When mass of all the molecules is halved and their speed is doubled, then the pressure will be

P′=13(m2)×NV×(2vrms)2P' = {1 \over 3}\left( {{m \over 2}} \right) \times {N \over V} \times {\left( {2{v_{rms}}} \right)^2}
=23mNVvrms2=2P= {2 \over 3}{{mN} \over V}v_{rms}^2 = 2P

(Using (i))

Q55
In the given (V −- T) diagram, what is the relation between pressure P 1 and P 2 ?
A P 2 < P 1
B Carnot be predicted
C P 2 = P 1
D P 2 > P 1
Correct Answer
Option A
Solution

According to ideal gas equation PV = nRT

⇒V=nRTP\Rightarrow V = {{nRT} \over P}

For an isobaric process, P = constant and V ∝\propto T Therefore, V – T graph is a straight line passing through origin.

Slope of this line is inversely proportional to P.

In the given figure, (Slope) 2 > (Slope) 1 ∴\therefore P 2 < P 1

Q56
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of CpCp{{{C_p}} \over {{C_p}}} for the gas is
A 53{5 \over 3}
B 32{3 \over 2}
C 43{4 \over 3}
D 2
Correct Answer
Option B
Solution

According to question

P∝T3P \propto {T^3}

But as we know for an adiabatic process the pressure

P∝Tγγ−1P \propto {T^{{\gamma \over {\gamma - 1}}}}

So,

γγ−1=3⇒γ=32{\gamma \over {\gamma - 1}} = 3 \Rightarrow \gamma = {3 \over 2}

∴\therefore

CpCv=32{{{C_p}} \over {{C_v}}} = {3 \over 2}
Q57
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 Zero
B −- 2000 J
C 2000 J
D 1000 J
Correct Answer
Option D
Solution

W net = Area of triangle ABC

=12AC×BC= {1 \over 2}AC \times BC
=12×5×10−3×4×105=1000 J= {1 \over 2} \times 5 \times {10^{ - 3}} \times 4 \times {10^5} = 1000\,J
Q58
The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T 1 K to T 2 K is
A 34NakB(T2−T1){3 \over 4}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)
B 34NakB(T2T1){3 \over 4}{N_a}{k_B}\left( {{{{T_2}} \over {{T_1}}}} \right)
C 38NakB(T2−T1){3 \over 8}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)
D 32NakB(T2−T1){3 \over 2}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)
Correct Answer
Option C
Solution

From first law of thermodynamics

ΔQ=ΔU+ΔW=32.14R(T2−T1)+0\Delta Q = \Delta U + \Delta W = {3 \over 2}.{1 \over 4}R\left( {{T_2} - {T_1}} \right) + 0
=38NaKB(T2−T1)= {3 \over 8}{N_a}{K_B}\left( {{T_2} - {T_1}} \right)
[∵K=RN]\left[\because{K = {R \over N}} \right]
Q59
An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. If Q 1 , Q 2 , Q 3 indicate the heat absorbed by the gas along the three processes and Δ\Delta U 1 , Δ\Delta U 2 , Δ\Delta U 3 indicate the change in internal energy along the three processes respectively, then
A Q 1 > Q 2 > Q 3 and Δ\Delta U 1 = Δ\Delta U 2 = Δ\Delta U 3
B Q 3 > Q 2 > Q 1 and Δ\Delta U 1 = Δ\Delta U 2 = Δ\Delta U 3
C Q 1 = Q 2 = Q 3 and Δ\Delta U 1 > Δ\Delta U 2 > Δ\Delta U 3
D Q 3 = Q 2 = Q 1 and Δ\Delta U 1 > Δ\Delta U 2 > Δ\Delta U 3
Correct Answer
Option A
Solution

Initial and final condition is same for all process

ΔU1=ΔU2=ΔU3\Delta {U_1} = \Delta {U_2} = \Delta {U_3}

from first law of thermodynamics

ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W

Work done

ΔW1>ΔW2>ΔW3\Delta {W_1} > \Delta {W_2} > \Delta {W_3}

(Area of P.V. graph) So

ΔQ1>ΔQ2>ΔQ3\Delta {Q_1} > \Delta {Q_2} > \Delta {Q_3}
Q60
A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is
A 2PV
B 4PV
C 12{1 \over 2} PV
D PV
Correct Answer
Option A
Solution

In a cyclic process,

ΔU=0\Delta U = 0

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

∴Δ\therefore \Delta

W = – Area of rectangle ABCD = – P(2V) = – 2PV According to first law of thermodynamics

ΔQ=ΔU+ΔW⇒ΔQ=ΔW\Delta Q = \Delta U + \Delta W \Rightarrow \Delta Q = \Delta W

(As

ΔU=0\Delta U = 0

) i.e., heat supplied to the system is equal to the work done So heat absorbed,

Δ\Delta

Q =

Δ\Delta

W = – 2PV ∴\therefore Heat rejected by the gas = 2PV

Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →