Heat and Thermodynamics — Page 10 | NEET Physics

Q91

Two Carnot engines A and B operate in series such that engine A absorbs heat at T1 and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at T3. When workdone in both the cases is equal, to value of T is :

A

{2 \over 3}{T_1} + {3 \over 2}{T_3}

B

{1 \over 3}{T_1} + {2 \over 3}{T_3}

C

{3 \over 2}{T_1} + {1 \over 3}{T_3}

D

{2 \over 3}{T_1} + {1 \over 3}{T_3}

Correct Answer

Option D

Solution

{W_A} = 1 - {{{Q_2}} \over {{Q_1}}} = 1 - {T \over {{T_1}}} \Rightarrow {{{Q_2}} \over {{Q_1}}} = {T \over {{T_1}}}

{W_B} = 1 - {{{Q_3}} \over {({Q_2}/2)}} = 1 - {{{T_3}} \over T} \Rightarrow {{2{Q_3}} \over {{Q_2}}} = {{{T_3}} \over T}

Now, WA = WB

{Q_1} - {Q_2} = {{{Q_2}} \over 2} - {Q_3}

\Rightarrow {{2{Q_1}} \over {{Q_2}}} + {{2{Q_3}} \over {{Q_2}}} = 3

\Rightarrow {{2{T_1}} \over T} + {{{T_3}} \over T} = 3

$\Rightarrow$

{{2{T_1}} \over 3} + {{{T_3}} \over 3} = T

Q92

A sample of gas at temperature

T

is adiabatically expanded to double its volume. The work done by the gas in the process is

\left(\mathrm{given}, \gamma=\dfrac{3}{2}\right)

:

A

W=T R[\sqrt{2}-2]

B

W=\dfrac{T}{R}[\sqrt{2}-2]

C

W=\dfrac{R}{T}[2-\sqrt{2}]

D

W=R T[2-\sqrt{2}]

Correct Answer

Option D

Solution

$\gamma=\dfrac{3}{2}$

\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)

\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}

Q93

The pressure

(\mathrm{P})

and temperature (

\mathrm{T})

relationship of an ideal gas obeys the equation

\mathrm{PT}^{2}=

constant. The volume expansion coefficient of the gas will be :

A

3 T^{2}

B

\dfrac{3}{T^2}

C

\dfrac{3}{T^3}

D

\dfrac{3}{T}

Correct Answer

Option D

Solution

PT^2

= constant From

PV = nRT \Rightarrow {{{T^3}} \over V} =

constant

{T^3} \propto V

..... (1)

3{T^2}dT \propto dV

..... (2) From (1) and (2)

{{3dT} \over T} = {{dV} \over V}

$\therefore$

\gamma = {1 \over V}{{dV} \over {dT}} = {3 \over T}

Q94

At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at

47^{\circ} \mathrm{C}

?

A 20 K

B 80 K

C 4 K

D

-73

K

Correct Answer

Option A

Solution

\begin{aligned} & \sqrt{\frac{3 R T}{2}}=\sqrt{\frac{3 R(320)}{32}} \\ & T=\frac{320}{16}=20 \mathrm{~K} \end{aligned}

Q95

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.Reason (R) : In isothermal process, PV = constant, while in adiabatic process

PV^{\gamma}

= constant. Here

\gamma

is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas.In the light of the above statements, choose the correct answer from the options given below:

A (A) is true but (R) is false

B Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

C (A) is false but (R) is true

D Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct Answer

Option D

Solution

We know, ideal gas equation, $\mathrm{Pv = nRT}$ For isothermal process, T = constant $\mathrm{\Rightarrow PV = const.}$ $\mathrm{\Rightarrow P \propto \dfrac{1}{V}}$ For an adiabatic process, $\mathrm{PV^r=constant}$ We can see, for the same pressure incresement,

{P_2} - {P_1}

,

\left( {{v_4} - {v_3}} \right) > \left( {{v_2} - {v_1}} \right)

i.e. the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.

Hence, option 4 is corret.

Q96

Consider the sound wave travelling in ideal gases of

\mathrm{He}, \mathrm{CH}_4

, and

\mathrm{CO}_2

. All the gases have the same ratio

\dfrac{P}{\rho}

, where

P

is the pressure and

\rho

is the density. The ratio of the speed of sound through the gases

\mathrm{V}_{\mathrm{He}}: \mathrm{V}_{\mathrm{CH}_4}: \mathrm{V}_{\mathrm{CO}_2}

is given by

A

\sqrt{\dfrac{7}{5}}: \sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{4}{3}}

B

\sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{4}{3}}: \sqrt{\dfrac{4}{3}}

C

\sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{4}{3}}: \sqrt{\dfrac{7}{5}}

D

\sqrt{\dfrac{4}{3}}: \sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{7}{5}}

Correct Answer

Option C

Solution

The speed of sound in an ideal gas is given by

v = \sqrt{\gamma \frac{P}{\rho}},

where: $\gamma$ is the ratio of specific heats,

P

is the pressure, and $\rho$ is the density. Since the problem states that

\frac{P}{\rho}

is the same for all the gases, the speed of sound in each gas is determined solely by the factor

\sqrt{\gamma}

. Let's determine the appropriate $\gamma$ for each gas: Helium (He): Helium is a monatomic gas. For a monatomic gas,

\gamma = \frac{5}{3}

. Therefore,

v_{\mathrm{He}} \propto \sqrt{\frac{5}{3}}

.

Methane (CH $_4$ ): Methane is a polyatomic gas (a tetrahedral molecule with three rotational degrees of freedom).

For nonlinear polyatomic gases, $\gamma$ is typically taken as

\frac{4}{3}

. Thus,

v_{\mathrm{CH_4}} \propto \sqrt{\frac{4}{3}}

. Carbon Dioxide (CO $_2$ ): Carbon dioxide is a linear molecule. For linear molecules,

\gamma = \frac{7}{5}

. Hence,

v_{\mathrm{CO_2}} \propto \sqrt{\frac{7}{5}}

. Therefore, the ratio of the speeds of sound in the gases is:

v_{\mathrm{He}} : v_{\mathrm{CH_4}} : v_{\mathrm{CO_2}} = \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{7}{5}}.

Comparing this expression with the given options, we see that it matches Option C.

Q97

An ideal gas exists in a state with pressure

P_0

, volume

V_0

. It is isothermally expanded to 4 times of its initial volume

\left(\mathrm{V}_0\right)

, then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is

A

\mathrm{P}_0 \mathrm{~V}_0(\ln 2-0.75)

B

\mathrm{P}_0 \mathrm{~V}_0(2 \ln 2-0.75)

C

\mathrm{P}_0 \mathrm{~V}_0(2 \ln 2-0.25)

D

\mathrm{P}_0 \mathrm{~V}_0(\ln 2-0.25)

Correct Answer

Option B

Solution

\begin{aligned} & \omega_1=\mathrm{P}_0 \mathrm{v}_0 \ell \mathrm{n} 4 \\ & \omega_2=\frac{\mathrm{P}_0}{4}\left(-3 \mathrm{v}_0\right)=-\frac{3 \mathrm{P}_0 \mathrm{v}_0}{4} \\ & \omega_3=0 \\ & \mathrm{Q}_{\mathrm{T}}=\Delta \mathrm{U}_{\text{cyclic }}+\omega \\ & \mathrm{Q}_{\mathrm{T}}=\omega \quad\left(\Delta \mathrm{U}_{\text{cyclic }}=0\right) \\ & \mathrm{Q}_{\mathrm{T}}=\mathrm{P}_0 \mathrm{v}_0\left(\ln 4-\frac{3}{4}\right) \\ & =\mathrm{P}_0 \mathrm{v}_0(2 \ln 2-0.75) \end{aligned}