Practice Start Test

Heat and Thermodynamics

NEET Physics · 97 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
A container of volume 200 cm3200 \mathrm{~cm}^3 contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature 200 K200 \mathrm{~K} (R=8.3JK−1 mol−1\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}) will be :-
A 6.15×105 Pa6.15 \times 10^5 \mathrm{~Pa}
B 6.15×104 Pa6.15 \times 10^4 \mathrm{~Pa}
C 4.15×105 Pa4.15 \times 10^5 \mathrm{~Pa}
D 4.15×106 Pa4.15 \times 10^6 \mathrm{~Pa}
Correct Answer
Option D
Solution
Pmix =(μ1+μ2)RTmix Vmix =(0.2+0.3)×8.3×200200×10−6=0.5×8.3×200200×10−6Pmix =4.15×106Pa\begin{aligned} & P_{\text{mix }}=\frac{\left(\mu_1+\mu_2\right) \mathrm{RT}_{\text{mix }}}{\mathrm{V}_{\text{mix }}} \\ & =\frac{(0.2+0.3) \times 8.3 \times 200}{200 \times 10^{-6}} \\ & =\frac{0.5 \times 8.3 \times 200}{200 \times 10^{-6}} \\ & \mathrm{P}_{\text{mix }}=4.15 \times 10^6 \mathrm{P}_{\mathrm{a}} \end{aligned}
Q12
The temperature of a gas is −50∘C-50^{\circ} \mathrm{C}. To what temperature the gas should be heated so that the rms speed is increased by 33 times?
A 3295∘C3295^{\circ} \mathrm{C}
B 3097 K3097 \mathrm{~K}
C 223 K223 \mathrm{~K}
D 669∘C669^{\circ} \mathrm{C}
Correct Answer
Option A
Solution
vrms∝Tv_{\mathrm{rms}} \propto \sqrt{T}
v1v2=T1 T2\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}

== let initial speed is

v\mathrm{v}

As speed is increased by

33

times so final speed become

4v4 v
⇒v4v=223 T\Rightarrow \frac{\mathrm{v}}{4 \mathrm{v}}=\sqrt{\frac{223}{\mathrm{~T}}}
T=3568 K\mathrm{T}=3568 \mathrm{~K}

So temp. in

∘C=3568−273=3295∘C{ }^{\circ} \mathrm{C}=3568-273=3295^{\circ} \mathrm{C}
Q13
A Carnot engine has an efficiency of 50%50 \% when its source is at a temperature 327∘C327^{\circ} \mathrm{C}. The temperature of the sink is :-
A 15∘C15^{\circ} \mathrm{C}
B 100∘C100^{\circ} \mathrm{C}
C 200∘C200^{\circ} \mathrm{C}
D 27∘C27^{\circ} \mathrm{C}
Correct Answer
Option D
Solution

Efficiency of carnot engine

%η=(1−Tsink Tsource )×100Tsource =327∘C=600 K50=(1−Tsink 600)×10012=1−Tsink 600 TSink =300 K\begin{aligned} & \% \eta=\left(1-\frac{T_{\text{sink }}}{\mathrm{T}_{\text{source }}}\right) \times 100 \\ & T_{\text{source }}=327^{\circ} \mathrm{C}=600 \mathrm{~K} \\ & 50=\left(1-\frac{\mathrm{T}_{\text{sink }}}{600}\right) \times 100 \\ & \frac{1}{2}=1-\frac{\mathrm{T}_{\text{sink }}}{600} \\ & \mathrm{~T}_{\text{Sink }}=300 \mathrm{~K} \end{aligned}

So temp. of sink is

∘C=300−2763=27∘C{ }^{\circ} \mathrm{C}=300-2763=27^{\circ} \mathrm{C}
Q14
An ideal gas follows a process described by the equation PV2=CP{V^2} = C from the initial (P1, V1, T1)({P_1},\,{V_1},\,{T_1}) to final (P2, V2, T2)({P_2},\,{V_2},\,{T_2}) thermodynamic states, where C is a constant. Then
A If P1>P2{P_1} > {P_2} then V1>V2{V_1} > {V_2}
B If P1>P2{P_1} > {P_2} then T1<T2{T_1} < {T_2}
C If V2>V1{V_2} > {V_1} then T2>T1{T_2} > {T_1}
D If V2>V1{V_2} > {V_1} then T2<T1{T_2} < {T_1}
Correct Answer
Option D
Solution

We know,

PV=nRTPV = nRT

Given,

PV2=P{V^2} =

constant ⇒\Rightarrow If

P1>P2⇒V1{P_1} > {P_2} \Rightarrow {V_1}

P{\left( {{{nRT} \over P}} \right)^2} =

constantconstant

{P^{ - 1}}{T^2} =

constantconstant

\Rightarrow P \propto {T^2}

i.e.,ifi.e., if

{P_1} > {P_2} \Rightarrow {T_1} > {T_2}

Also,Also,

\left( {{{nRT} \over V}} \right){V^2} =

constantconstant

TV =

constantconstant

\Rightarrow

IfIf

{V_2} > {V_1}

thenthen

{T_2}

Q15
Two rods one made of copper and other made of steel of same length and same cross sectional area are joined together. The thermal conductivity of copper and steel are 385 J s −-1 K −-1 m −-1 and 50 J s −-1 K −-1 m −-1 respectively. The free ends of copper and steel are held at 100∘^\circC and 0∘^\circC respectively. The temperature at the junction is, nearly :
A 88.5∘C88.5^\circ C
B 12∘C12^\circ C
C 50∘C50^\circ C
D 73∘C73^\circ C
Correct Answer
Option A
Solution

We know in conduction, rate of flow of heat

q=−K.A.ΔTΔxq = {{ - K.A.\Delta T} \over {\Delta x}}

As it is a case of steady state heat transfer

⇒−Kcu×A×[100−Tj]l=−Ksteel×A×[Tj−0]l\Rightarrow {{ - {K_{cu}} \times A \times [100 - {T_j}]} \over l} = {{ - {K_{steel}} \times A \times [{T_j} - 0]} \over l}

Where T j is temperature of the junction

⇒385(100−Tj)=50(Tj)\Rightarrow 385(100 - {T_j}) = 50({T_j})
⇒100Tj−1=50385⇒100Tj=1+50385⇒100Tj=1.1298\Rightarrow {{100} \over {{T_j}}} - 1 = {{50} \over {385}} \Rightarrow {{100} \over {{T_j}}} = 1 + {{50} \over {385}} \Rightarrow {{100} \over {{T_j}}} = 1.1298
⇒Tj=88.5∘C\Rightarrow {T_j} = 88.5^\circ C
Q16
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains helium (monoatomic), the second contains fluorine (diatomic) and the third contains sulfur hexafluoride (polyatomic). The correct statement, among the following is :
A The root mean square speed of sulfur hexafluoride is the largest
B All vessels contain unequal number of respective molecules
C The root mean square speed of molecules is same in all three cases
D The root mean square speed of helium is the largest
Correct Answer
Option D
Solution

All three vessels have equal volume and same temperature and pressure. From ideal gas equation

PV=nRTPV = nRT
nR=PVTnR = {{PV} \over T}
n=PVRT=n = {{PV} \over {RT}} =

constant So, here all three vessels contains equal number of moles and number of gas molecules. Now,

vrms=3RTM⇒vrms∝1M{v_{rms}} = \sqrt {{{3RT} \over M}} \Rightarrow {v_{rms}} \propto {1 \over {\sqrt M }}

Here, rms speed of helium is the largest.

Q17
An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is
A 1
B 2
C 3
D 4
Correct Answer
Option B
Solution
(dPdV)adiabatic=−γP{\left( {{{dP} \over {dV}}} \right)_{adiabatic}} = - \gamma P
(dPdV)isothermal=−P{\left( {{{dP} \over {dV}}} \right)_{isothermal}} = -P
(dPdV)adiabatic>(dPdV)isothermal{\left( {{{dP} \over {dV}}} \right)_{adiabatic}} > {\left( {{{dP} \over {dV}}} \right)_{isothermal}}
Q18
The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is
A 5.6 ×\times 10 6 m 3
B 5.6 ×\times 10 3 m 3
C 5.6 ×\times 10 −-3 m 3
D 5.6 m 3
Correct Answer
Option D
Solution

From ideal gas equation

PV=nRTPV = nRT
[n=mass of watermol. wt.=4.5×10318]\left[ {n = {{mass\,of\,water} \over {mol.\,wt.}} = {{4.5 \times {{10}^3}} \over {18}}} \right]
V=nRTPV = {{nRT} \over P}

At

STP⇒T=273 KSTP \Rightarrow T = 273\,K
P=105P = {10^5}

N/m 2

V=4.5×10318×8.3×273105=5.66V = {{4.5 \times {{10}^3}} \over {18}} \times {{8.3 \times 273} \over {{{10}^5}}} = 5.66

m 3

Q19
A cup of coffee cools from 90∘^\circC to 80∘^\circC in t minutes, when the room temperature is 20∘^\circC. The time taken by a similar cup of coffee to cool from 80∘^\circC to 60∘^\circC at a room temperature same at 20∘^\circC is :
A 513{5 \over {13}}t
B 1310{13 \over {10}}t
C 135{13 \over {5}}t
D 1013{10 \over {13}}t
Correct Answer
Option C
Solution
dTdt=K(Tav−T0){{dT} \over {dt}} = K({T_{av}} - {T_0})
10t=K(85−20){{10} \over t} = K(85 - 20)
20t=K(70−20){{20} \over t} = K(70 - 20)
t′2t=6550{{t'} \over {2t}} = {{65} \over {50}}
t′t=13050{{t'} \over t} = {{130} \over {50}}
t′=135tt' = {{13} \over 5}t
Q20
Match Column - I and Column - II and choose the correct match from the given choices.<br><br><table> <thead> <tr> <th></th> <th>Column - I</th> <th></th> <th>Column - II</th> </tr> </thead> <tbody> <tr> <td>(A)</td> <td>Root mean square speed of gas molecules</td> <td>(P)</td> <td>13nmv‾2{1 \over 3}nm{\overline v ^2}</td> </tr> <tr> <td>(B)</td> <td>Pressure exerted by ideal gas</td> <td>(Q)</td> <td>3RTM\sqrt {{{3RT} \over M}} </td> </tr> <tr> <td>(C)</td> <td>Average kinetic energy of a molecule</td> <td>(R)</td> <td>52RT{5 \over 2}RT</td> </tr> <tr> <td>(D)</td> <td>Total internal energy of 1 mole of a diatomic gas</td> <td>(S)</td> <td>32kBT{3 \over 2}{k_B}T</td> </tr> </tbody> </table>
A (A) - (R), (B) - (Q), (C) - (P), (D) - (S)
B (A) - (R), (B) - (P), (C) - (S), (D) - (Q)
C (A) - (Q), (B) - (R), (C) - (S), (D) - (P)
D (A) - (Q), (B) - (P), (C) - (S), (D) - (R)
Correct Answer
Option D
Solution

(A)

Vrms=3RTM{V_{rms}} = \sqrt {{{3RT} \over M}}

(B)

P=P =
13nmv‾2{1 \over 3}nm{\overline v ^2}

(C)

E=32kTE = {3 \over 2}kT

(D)

ETotal=nf2RT=52RT{E_{Total}} = n{f \over 2}RT = {5 \over 2}RT
Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →