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Heat and Thermodynamics

JEE Physics · 315 questions · Page 19 of 32 · Click an option or "Show Solution" to reveal answer

Q181
A Carnot engine has efficiency of 50%50 \%. If the temperature of sink is reduced by 40∘C40^{\circ} \mathrm{C}, its efficiency increases by 30%30 \%. The temperature of the source will be:
A 166.7 K
B 255.1 K
C 266.7 K
D 367.7 K
Correct Answer
Option C
Solution
1−TLTH=0.51 - {{{T_L}} \over {{T_H}}} = 0.5

...... (1)

1−TL−40TH=0.651 - {{{T_L} - 40} \over {{T_H}}} = 0.65

.... (2)

⇒TH=8003 K≃266.7 K\Rightarrow {T_H} = {{800} \over 3}\,K \simeq 266.7\,K
Q182
Given below are two statements : Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature. Statement II : The rms speed of oxygen molecules in a gas is vv. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become 2v2 v. In the light of the above statements, 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 D
Solution

Average momentum

=⟨P→⟩=0= \left\langle {\overrightarrow P } \right\rangle = 0
vrms=3RTM{v_{rms}} = \sqrt {{{3RT} \over M}}

If temperature is doubled and oxygen atoms are used then

vrms′=3R(2T)M/2=4 vrmsv{'_{rms}} = \sqrt {{{3R(2T)} \over {M/2}}} = 4\,{v_{rms}}
Q183
The root mean square speed of smoke particles of mass 5×10−17 kg5 \times 10^{-17} \mathrm{~kg} in their Brownian motion in air at NTP is approximately. [Given k=1.38×10−23JK−1\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}]
A 60 mm s−160 \mathrm{~mm} \mathrm{~s}^{-1}
B 12 mm s−112 \mathrm{~mm} \mathrm{~s}^{-1}
C 15 mm s−115 \mathrm{~mm} \mathrm{~s}^{-1}
D 36 mm s−136 \mathrm{~mm} \mathrm{~s}^{-1}
Correct Answer
Option C
Solution

At NTP, T=298 KT=298 \mathrm{~K}

vrms=3RTM=3kNA×2985×10−17×NA\begin{aligned} v_{\mathrm{rms}} &=\sqrt{\frac{3 R T}{M}} \\ &=\sqrt{\frac{3 k N_{A} \times 298}{5 \times 10^{-17} \times N_{A}}} \end{aligned}

≃15 mm/s\simeq 15 \mathrm{~mm} / \mathrm{s}

Q184
A Carnot engine operating between two reservoirs has efficiency 13\dfrac{1}{3}. When the temperature of cold reservoir raised by x, its efficiency decreases to 16\dfrac{1}{6}. The value of x, if the temperature of hot reservoir is 99∘99^\circC, will be :
A 66 K
B 62 K
C 16.5 K
D 33 K
Correct Answer
Option B
Solution

Given η=13\eta=\dfrac{1}{3} When T2\mathrm{T}_2 raised by x (T2\mathrm{T}_2 = (T2+x)\left(\mathrm{T}_2+\mathrm{x}\right)), its efficiency decreases to

16\frac{1}{6}

η′=16\eta^{\prime}=\dfrac{1}{6} Temperature of hot reservior (T1)=99∘C\left(\mathrm{T}_1\right)=99^{\circ} \mathrm{C}

=99+273=372∘K=99+273=372^{\circ} \mathrm{K}

As we know, efficiency of carnot engine

η=1−T2T1=13.....(1)η′=1−(T2+x)T1=16.....(2)η′=T1−(T2+x)T1=16\begin{aligned} & \eta=1-\frac{T_2}{T_1}=\frac{1}{3} .....(1) \\\\ & \eta^{\prime}=1-\frac{\left(T_2+x\right)}{T_1}=\frac{1}{6} .....(2) \\\\ & \eta^{\prime}=\frac{T_1-\left(T_2+x\right)}{T_1}=\frac{1}{6} \end{aligned}

From equation (1)

13=1−T2372\frac{1}{3}=1-\frac{\mathrm{T}_2}{372}
13=372−T2372⇒372−3723=T2⇒ T2=248 K\begin{aligned} & \frac{1}{3}=\frac{372-\mathrm{T}_2}{372} \\\\ &\Rightarrow 372-\frac{372}{3}=\mathrm{T}_2 \\\\ &\Rightarrow \mathrm{~T}_2=248 \mathrm{~K} \end{aligned}

By putting the value of T2\mathrm{T}_2 in equation (2)

T1−(T2+x)T1=16⇒372−(248+x)372=16⇒372−24−x=3726⇒124−x=62⇒124−62=x⇒x=62 K\begin{aligned} & \frac{T_1-\left(T_2+x\right)}{T_1}=\frac{1}{6} \\\\ &\Rightarrow \frac{372-(248+x)}{372}=\frac{1}{6} \\\\ &\Rightarrow 372-24-x=\frac{372}{6} \\\\ &\Rightarrow 124-x=62 \\\\ &\Rightarrow 124-62=x \\\\ &\Rightarrow x=62 \mathrm{~K} \end{aligned}
Q185
Heat energy of 735 J735 \mathrm{~J} is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be :
A 572 J572 \mathrm{~J}
B 441 J441 \mathrm{~J}
C 525 J525 \mathrm{~J}
D 735 J735 \mathrm{~J}
Correct Answer
Option C
Solution

ΔQ=nCPΔT=\Delta Q=n C_{P} \Delta T=

n(f2+1)RΔTn\left( {{f \over 2} + 1} \right)R\Delta T

=

n(52+1)RΔTn\left( {{5 \over 2} + 1} \right)R\Delta T

=

n(72)RΔTn\left( {{7 \over 2}} \right)R\Delta T

Given, ΔQ\Delta Q = 735 ⇒\Rightarrow

n(72)RΔTn\left( {{7 \over 2}} \right)R\Delta T

= 735 ⇒\Rightarrow

nRΔT=735×27nR\Delta T = 735 \times {2 \over 7}

Also,

Δ\Delta

U =

f2nRΔT{f \over 2}nR\Delta T

=

52nRΔT{5 \over 2}nR\Delta T

=

52×735×27{5 \over 2} \times 735 \times {2 \over 7}

= 525 J

Q186
A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is 1681\dfrac{16}{81}. Then the ratio of CpCv\dfrac{\mathrm{Cp}}{\mathrm{Cv}} will be.
A 31\dfrac{3}{1}
B 43\dfrac{4}{3}
C 12\dfrac{1}{2}
D 32\dfrac{3}{2}
Correct Answer
Option B
Solution

Let γ\gamma be the ratio of CpCv\dfrac{C_{p}}{C_{v}} Then for adiabatic process

PVγ= Constant ⇒PiPf=(VfVi)γ⇒8116=(278)γ⇒γ=43\begin{aligned} & P V^{\gamma}=\text{ Constant } \\\\ & \Rightarrow \frac{P_{i}}{P_{f}}=\left(\frac{V_{f}}{V_{i}}\right)^{\gamma} \\\\ & \Rightarrow \frac{81}{16}=\left(\frac{27}{8}\right)^{\gamma} \\\\ & \Rightarrow \gamma=\frac{4}{3} \end{aligned}
Q187
A sample of gas at temperature TT is adiabatically expanded to double its volume. The work done by the gas in the process is (given,γ=32)\left(\mathrm{given}, \gamma=\dfrac{3}{2}\right) :
A W=TR[2−2]W=T R[\sqrt{2}-2]
B W=TR[2−2]W=\dfrac{T}{R}[\sqrt{2}-2]
C W=RT[2−2]W=\dfrac{R}{T}[2-\sqrt{2}]
D W=RT[2−2]W=R T[2-\sqrt{2}]
Correct Answer
Option D
Solution

γ=32\gamma=\dfrac{3}{2}

W=nRΔT1−γ=nRTf−nRTi1−γ=(PV)f−(PVi)1−γ… (1) PVγ= constant PiViγ=Pf(2Vi)γ⇒Pf=Pi2γ=Pi22......(2)\begin{aligned} & W =\frac{n R \Delta T}{1-\gamma}=\frac{n R T_{f}-n R T_{i}}{1-\gamma} \\\\ & =\frac{(P V)_{f}-\left(P V_{i}\right)}{1-\gamma} \quad \ldots \text{ (1) } \\\\ & P V^{\gamma}=\text{ constant } \\\\ & P_{i} V_{i}^{\gamma}=P_{f}\left(2 V_{i}\right)^{\gamma} \Rightarrow P_{f}=\frac{P_{i}}{2^{\gamma}}=\frac{P_{i}}{2 \sqrt{2}} ......(2) \end{aligned}

From (1) and (2)

W=Pi222Vi−PiVi1−γ=PiVi−1/2(12−1)=−nRT(2−2)=nRT(2−2)\begin{aligned} & W=\frac{\frac{P_{i}}{2 \sqrt{2}} 2 V_{i}-P_{i} V_{i}}{1-\gamma}=\frac{P_{i} V_{i}}{-1 / 2}\left(\frac{1}{\sqrt{2}}-1\right) \\\\ & =-n R T(\sqrt{2}-2) \\\\ & =n R T(2-\sqrt{2}) \end{aligned}
Q188
(P+aV2)(V−b)=RT\left(P+\dfrac{a}{V^{2}}\right)(V-b)=R T represents the equation of state of some gases. Where PP is the pressure, VV is the volume, TT is the temperature and a,b,Ra, b, R are the constants. The physical quantity, which has dimensional formula as that of b2a\dfrac{b^{2}}{a}, will be:
A Energy density
B Bulk modulus
C Modulus of rigidity
D Compressibility
Correct Answer
Option D
Solution

[a]=[ML5 T−2][a]=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]

[b]=[L3][b2a]=[L6ML5 T2]=[M−1LT−2]=[ Compressibility] \begin{aligned} & {[b]=\left[\mathrm{L}^{3}\right] } \\\\ & {\left[\frac{b^{2}}{a}\right]=\left[\frac{\mathrm{L}^{6}}{\mathrm{ML}^{5} \mathrm{~T}^{2}}\right] }=\left[\mathrm{M}^{-1} \mathrm{LT}^{-2}\right] \\\\ &=[\text{ Compressibility] } \end{aligned}
Q189
The average kinetic energy of a molecule of the gas is
A proportional to volume
B dependent on the nature of the gas
C proportional to absolute temperature
D proportional to pressure
Correct Answer
Option C
Solution

Average kinetic energy of a molecule of gas

=f2kBT=\frac{f}{2} k_{B} T

f is degree of freedom.

Q190
The correct relation between γ=cpcv\gamma = {{{c_p}} \over {{c_v}}} and temperature T is :
A γ∝T\gamma \propto T
B γ∝1T\gamma \propto {1 \over {\sqrt T }}
C γ∝1T\gamma \propto {1 \over T}
D γ∝T∘\gamma \propto T^\circ
Correct Answer
Option D
Solution

γ=CPCV\gamma=\dfrac{C_{P}}{C_{V}} At low temperature (T),γ(T), \gamma is independent of TT.

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