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Heat and Thermodynamics

JEE Physics · 315 questions · Page 16 of 32 · Click an option or "Show Solution" to reveal answer

Q151
The height of victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall? [Given 1 cal = 4.2 J and specific heat of water = 1 cal g−-1 ∘^\circ0C−-1]
A 0.147∘^\circ C
B 14.76∘^\circ C
C 1.476∘^\circ C
D 0.014∘^\circ C
Correct Answer
Option A
Solution

Change in P.E. = Heat energy mgh = mS

Δ\Delta

T

Δ\Delta

T =

ghS{{gh} \over S}

=

10×634200J/kgC{{10 \times 63} \over {4200J/kgC}}

= 0.147

∘^\circ

C

Q152
A reversible engine has an efficiency of 14{1 \over 4}. If the temperature of the sink is reduced by 58∘^\circC, its efficiency becomes double. Calculate the temperature of the sink :
A 174∘^\circC
B 280∘^\circC
C 180.4∘^\circC
D 382∘^\circC
Correct Answer
Option A
Solution

T2 = sink temperature

η=1−T2T1\eta = 1 - {{{T_2}} \over {{T_1}}}
14=1−T2T1{1 \over 4} = 1 - {{{T_2}} \over {{T_1}}}
T2T1=34{{{T_2}} \over {{T_1}}} = {3 \over 4}

.... (i)

12=1−T2−58T1{1 \over 2} = 1 - {{{T_2} - 58} \over {{T_1}}}
T2T1=58T1=12{{{T_2}} \over {{T_1}}} = {{58} \over {{T_1}}} = {1 \over 2}
34=58T1+12{3 \over 4} = {{58} \over {{T_1}}} + {1 \over 2}
14=58T1⇒T1=232{1 \over 4} = {{58} \over {{T_1}}} \Rightarrow {T_1} = 232
T2=34×232{T_2} = {3 \over 4} \times 232
T2=174{T_2} = 174

K

Q153
For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation dpdv=−ap{{dp} \over {dv}} = - ap. If p = p0 at v =0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)
A p0aeR{{{p_0}} \over {aeR}}
B ap0eR{{a{p_0}} \over {eR}}
C infinity
D 0∘^\circC
Correct Answer
Option A
Solution
∫p0pdpP=−a∫0vdv\int\limits_{{p_0}}^p {{{dp} \over P} = - a\int\limits_0^v {dv} }
ln⁡(pp0)=−av\ln \left( {{p \over {{p_0}}}} \right) = - av
p=p0e−avp = {p_0}{e^{ - av}}

For temperature maximum p-v product should be maximum

T=pvnR=p0ve−avRT = {{pv} \over {nR}} = {{{p_0}v{e^{ - av}}} \over R}
dTdv=0⇒p0R{e−av+ve−av(−a)}{{dT} \over {dv}} = 0 \Rightarrow {{{p_0}} \over R}\{ {e^{ - av}} + v{e^{ - av}}( - a)\}

= 0

p0e−avR{1−av}=0{{{p_0}{e^{ - av}}} \over R}\{ 1 - av\} = 0
v=1a,∞v = {1 \over a},\infty
T=p01Rae=p0RaeT = {{{p_0}1} \over {Rae}} = {{{p_0}} \over {Rae}}

at v = ∞\infty T = 0 Option (a)

Q154
Two thin metallic spherical shells of radii r1 and r2 (r1 < r2) are placed with their centres coinciding. A material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature θ\theta1 and the outer shell at temperature θ\theta2(θ\theta1 < θ\theta2). The rate at which heat flows radially through the material is :-
A 4πKr1r2(θ2−θ1)r2−r1{{4\pi K{r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}
B πr1r2(θ2−θ1)r2−r1{{\pi {r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}
C K(θ2−θ1)r2−r1{{K({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}
D K(θ2−θ1)(r2−r1)4πr1r2{{K({\theta _2} - {\theta _1})({r_2} - {r_1})} \over {4\pi {r_1}{r_2}}}
Correct Answer
Option A
Solution

Thermal resistance of spherical sheet of thickness dr and radius r is

dR=drK(4πr2)dR = {{dr} \over {K(4\pi {r^2})}}
R=∫r1r2drK(4πr2)R = \int\limits_{{r_1}}^{{r_2}} {{{dr} \over {K(4\pi {r^2})}}}
R=14πK(1r1−1r2)=14πK(r2−r1r1r2)R = {1 \over {4\pi K}}\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right) = {1 \over {4\pi K}}\left( {{{{r_2} - {r_1}} \over {{r_1}{r_2}}}} \right)

Thermal current (i)

=θ2−θ1R= {{{\theta _2} - {\theta _1}} \over R}
i=4πKr1r2r2−r1(θ2−θ1)i = {{4\pi K{r_1}{r_2}} \over {{r_2} - {r_1}}}({\theta _2} - {\theta _1})
Q155
At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm ? Both the diameters have been measured at room temperature (27∘^\circC). (Given : coefficient of linear thermal expansion of gold α\alphaL = 1.4 ×\times 10−-5 K−-1)
A 125.7∘^\circC
B 91.7∘^\circC
C 425.7∘^\circC
D 152.7∘^\circC
Correct Answer
Option D
Solution
ΔD=DαΔT\Delta D = D\alpha \Delta T
ΔT=0.0116.230×1.4×10−5\Delta T = {{0.011} \over {6.230 \times 1.4 \times {{10}^{ - 5}}}}

= 126.11

∘^\circ

C ⇒\Rightarrow Tf = T +

Δ\Delta

T = (27 + 126.11)

∘^\circ

C = 153.11

∘^\circ

C

Q156
Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2, if the process is purely adiabatic and W3 if the process is purely isobaric. Then, choose the correct option
A W1 2 3
B W2 3 1
C W3 1 2
D W2 1 3
Correct Answer
Option D
Solution

Comparing the area under the PV graph A3 > A1 > A2 ⇒\Rightarrow W3 > W1 > W2

Q157
A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm3 is :
A 72 ×\times 105
B 32 ×\times 105
C 27 ×\times 104
D 54 ×\times 104
Correct Answer
Option C
Solution

Total number of moles are

n=nH2+nO2n = {n_{{H_2}}} + {n_{{O_2}}}
=162+12832= {{16} \over 2} + {{128} \over {32}}

= 12 moles Using

PV=nRTPV = nRT
V=nRTPV = {{nRT} \over P}
=12×8.31×273.15105= {{12 \times 8.31 \times 273.15} \over {{{10}^5}}}

m3 = 0.27 m3 = 27 ×\times 104 cm3

Q158
What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?
A The velocity of atomic oxygen remains same
B The velocity of atomic oxygen doubles
C The velocity of atomic oxygen becomes half
D The velocity of atomic oxygen becomes four times
Correct Answer
Option B
Solution

As

vrms=3RTM0{v_{rms}} = \sqrt {{{3RT} \over {{M_0}}}}

T is doubled and oxygen molecule is dissociated into atomic oxygen molar mass is halved. So,

vrms′=3RT×2T0M0/2=2vrmsv{'_{rms}} = \sqrt {{{3RT \times 2{T_0}} \over {{M_0}/2}}} = 2{v_{rms}}

So velocity of atomic oxygen is doubled.

Q159
Given below are two statements : Statement I : When μ\mu amount of an ideal gas undergoes adiabatic change from state (P1, V1, T1) to state (P2, V2, T2), then work done is W=μR(T2−T1)1−γW = {{\mu R({T_2} - {T_1})} \over {1 - \gamma }}, where γ=CpCv\gamma = {{{C_p}} \over {{C_v}}} and R = universal gas constant. Statement II : In the above case, when work is done on the gas, the temperature of the gas would rise. Choose the correct answer from the options given below :
A Both Statement I and Statement II are true.
B Both Statement I and Statement II are false.
C Statement I is true but Statement II is false.
D Statement I is false but Statement II is true.
Correct Answer
Option A
Solution
W=μR(T2−T1)1−rW = {{\mu R({T_2} - {T_1})} \over {1 - r}}

for a polytropic process for adiabatic process r = γ\gamma ⇒\Rightarrow Statement I is true. In an adiabatic process

Δ\Delta

U = −-

Δ\Delta

W ⇒\Rightarrow If work is done on the gas ⇒\Rightarrow

Δ\Delta

W is negative ⇒\Rightarrow

Δ\Delta

U is positive or temperature increases ⇒\Rightarrow Statement II is true

Q160
According to kinetic theory of gases, A. The motion of the gas molecules freezes at 0∘^\circC. B. The mean free path of gas molecules decreases if the density of molecules is increased. C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant. D. Average kinetic energy per molecule per degree of freedom is 32kBT{3 \over 2}{k_B}T (for monoatomic gases). Choose the most appropriate answer from the options given below :
A A and C only
B B and C only
C A and B only
D C and D only
Correct Answer
Option B
Solution

According to kinetic theory of gases, A.

The motion of the gas molecules freezes at 0 K.

B.

The mean free path decreases on increasing the number density of the molecules as

μ=12πnd2⇒μ∝1n\mu = {1 \over {\sqrt 2 \pi n{d^2}}} \Rightarrow \mu \propto {1 \over n}

.

C.

The mean free path increases on increasing the volume.

Now if temperature is increased by keeping the pressure constant the volume should increase that is mean free path increases.

D.

K.E.avg per molecule per degree of freedom is

12kBT{1 \over 2}{k_B}T

.

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