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Heat and Thermodynamics

NEET Physics · 97 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
A mass of diatomic gas (γ=1.4)(\gamma = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27 o C to 927 o C. The pressure of the gas in the final state is
A 8 atm
B 28 atm
C 68.7 atm
D 256 atm
Correct Answer
Option D
Solution

T 1 = 273 + 27 = 300K T 2 = 273 + 927 = 1200K For adiabatic process,

P1−γTγ{P^{1 - \gamma }}{T^\gamma }

= constant

⇒P11−γT1γ=P21−γT2γ\Rightarrow {P_1}^{1 - \gamma }{T_1}^\gamma = {P_2}^{1 - \gamma }{T_2}^\gamma
⇒(P2P1)1−γ=(T1T2)γ\Rightarrow {\left( {{{{P_2}} \over {{P_1}}}} \right)^{1 - \gamma }} = {\left( {{{{T_1}} \over {{T_2}}}} \right)^\gamma }
⇒(P1T2)1−γ=(T2T1)γ\Rightarrow {\left( {{{{P_1}} \over {{T_2}}}} \right)^{1 - \gamma }} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^\gamma }
(P1P2)1−1.4=(1200300)1.4{\left( {{{{P_1}} \over {{P_2}}}} \right)^{1 - 1.4}} = {\left( {{{1200} \over {300}}} \right)^{1.4}}
(P1P2)−0.4=(4)1.4{\left( {{{{P_1}} \over {{P_2}}}} \right)^{ - 0.4}} = {\left( 4 \right)^{1.4}}
(P2P1)0.4=41.4{\left( {{{{P_2}} \over {{P_1}}}} \right)^{0.4}} = {4^{1.4}}
P2=P14(1.40.4)=P14(72){P_2} = {P_1}{4^{\left( {{{1.4} \over {0.4}}} \right)}} = {P_1}{4^{\left( {{7 \over 2}} \right)}}

= P 1 (2 7 ) = 2 × 128 = 256 atm

Q62
During an isothermal expansion, a confined ideal gas does −- 150 J of work against its surroundings. This implies that
A 150 J of heat has been removed from the gas
B 300 J of heat has been added to the gas
C no heat is transferred because the process is isothermal
D 150 J of heat has been added to the gas
Correct Answer
Option D
Solution

If a process is expansion then work done is positive so answer will be (a).

But in question work done by gas is given –150J so that according to it answer will be (d).

Q63
When 1 kg of ice at 0 o C melts to water at 0 o C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/ o C, is
A 273 cal/K
B 8 × \times 10 4 cal/K
C 80 cal/K
D 293 cal/K
Correct Answer
Option D
Solution

Change in entropy is given by

dS=dQT⇒ΔS=ΔQT=mLf273dS = {{dQ} \over T} \Rightarrow \Delta S = {{\Delta Q} \over T} = {{m{L_f}} \over {273}}
ΔS=1000×80273=293 cal/K\Delta S = {{1000 \times 80} \over {273}} = 293\,cal/K

Note : In the question paper unit of latent heat of ice is given to be cal/°C.

It is wrong.

The unit of latent heat of ice is cal/g.

Q64
If c p and c v denote the specific heats (per unit mass of an ideal gas of molecular weight M, then
A c p −- c v = R/M 2
B c p −- c v = R
C c p −- c v = R/M
D c p −- c v = MR
Correct Answer
Option C
Solution

C v = molar specific heat of the ideal gas at constant volume C p = molar specific heat of the ideal gas at constant pressure, C p ' = MC p and C v ’ = MC v Also

Cp′−Cv′=R{C_p}' - {C_v}' = R

MC p – MC v = R C p – C v = R/M

Q65
A monatomic gas at pressure P 1 and volume V 1 is compressed adiabatically to 18th{{1 \over 8}^{th}} of its original volume. What is the final pressure of the gas?
A 64P 1
B P 1
C 16P 1
D 32P 1
Correct Answer
Option D
Solution

Ideal gas equation, for an adiabatic process is

PVγ=P{V^\gamma } =

constant ⇒\Rightarrow

P1V1γ=P2V2γ{P_1}V_1^\gamma = {P_2}V_2^\gamma

For monoatomic gas

γ=53\gamma = {5 \over 3}

∴\therefore

P1V15/3=P2(V18)5/3{P_1}V_1^{5/3} = {P_2}{\left( {{{{V_1}} \over 8}} \right)^{5/3}}

⇒\Rightarrow P 2 = P 1 × (2) 5 = 32 P 1 .

Q66
If Δ\Delta U and Δ\Delta W represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true ?
A Δ\Delta U = −-Δ\Delta W, in an adiabtic process
B Δ\Delta U = Δ\Delta W, in an isothermal process
C Δ\Delta U = Δ\Delta W, in an adiabatic process
D Δ\Delta U = −- Δ\Delta W, in an isothermal process
Correct Answer
Option A
Solution

By first law of thermodynamics,

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

In adiabatic process,

ΔQ=0\Delta Q = 0

∴\therefore

ΔU=−ΔW\Delta U = - \Delta W

In isothermal process,

ΔU=0\Delta U = 0

∴\therefore

ΔQ=ΔW\Delta Q = \Delta W
Q67
In thermodynamic processes which of the following statements is not true ?
A In an isochoric process pressure remains constant.
B In an isothermal process the temperature remains constant.
C In an adiabatic process PV γ\gamma = constant.
D In an adiabatic process the system is insulated from the surroundings.
Correct Answer
Option A
Solution

The only statement which is not true is statement (a).

A process in which the pressure remains constant is called isobaric process and not isochoric as in isochoric process the volume remains constant.

Q68
The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is
A 6400 J
B 5400 J
C 7900 J
D 8900 J
Correct Answer
Option C
Solution

When a quantity Q of heat is supplied to a system it is used to do an amount of work W by the system and to increase the internal energy of the system by ∆U : Q = ∆U + W Here, Q given in kilo calories is converted into joule.

Therefore Q = 2×1000×4.2 J = 8400 J The increase the internal energy ∆U = Q – W = 8400 – 500 = 7900 J

Q69
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then
A E = 0
B Q = 0
C W = 0
D Q = W = 0
Correct Answer
Option A
Solution

Internal energy depends only on the initial and final states of temperature and not on the path.

In a cyclic process, as initial and final states are the same, change in internal energy is zero.

Hence E is

Δ\Delta

U, the change in internal energy.

Q70
At 10 o C the value of the density of a fixed mass of an ideal gas divided by it pressure is x. At 110 o C this ratio is
A 10110x{{10} \over {110}}x
B 283383x{{283} \over {383}}x
C xx
D 383283x{{383} \over {283}}x
Correct Answer
Option B
Solution

PV = nRT

⇒PV=mMRT\Rightarrow PV = {m \over M}RT
⇒PVm=RTM\Rightarrow {{PV} \over m} = {{RT} \over M}
⇒Pρ=RTM\Rightarrow {P \over \rho } = {{RT} \over M}
⇒Pρ∝T\Rightarrow {P \over \rho } \propto T
∴Pρ∝1T\therefore {P \over \rho } \propto {1 \over T}
P1ρ1P2ρ2=T1T2=383283{{{{{P_1}} \over {{\rho _1}}}} \over {{{{P_2}} \over {{\rho _2}}}}} = {{{T_1}} \over {{T_2}}} = {{383} \over {283}}
⇒xP2P2=383283\Rightarrow {x \over {{{{P_2}} \over {{P_2}}}}} = {{383} \over {283}}
∴P2P2=383283x\therefore {{{P_2}} \over {{P_2}}} = {{383} \over {283}}x
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