Physics Part Ii Chapter 12 Thermodynamics

Work done during isothermal expansion of 1 mole of ideal gas

W = ∫ dW

v2
W = ∫ Pd V - - - - ( I )
v1

For 1 mole of gas

PV = RT

P = RT / V

Putting value of P in eqn ( I ).,

v2
W = ∫ RT / V dV
v1

v2
W = RT ∫ 1 / V dV
v1
v2
W= RT [ log e V]
v1

W = RT [ log e V2 - log e V1 ]

W = 2.303 RT log 10 V2 / V1

For constant temperature,

W = 2.303 RT log 10 P2 / P1

The second law of thermodynamics states that the total entropy can never decrease over time for an isolated system, that is, a system in which neither energy nor matter can enter nor leave. The total entropy can remain constant in ideal cases where the system is in a steady state (equilibrium), or is undergoing a reversible process. In all spontaneous processes, the total entropy always increases and the process is irreversible. The increase in entropy accounts for the irreversibility of natural processes[according to whom?, and the asymmetry between future and past.

Historically, the second law was an empirical finding that was accepted as an axiom of thermodynamic theory. Statistical thermodynamics, classical or quantum, explains the microscopic origin of the law.

The second law has been expressed in many ways. Its first formulation is credited to the French scientist Sadi Carnot, who in 1824 showed that there is an upper limit to the efficiency of conversion of heat to work, in a heat engine.

D.

2.24 × 10^–16 J

W = QdV = Q(Vq - VP)
= -100 × (1.6 × 10^-19) × (– 4 – 10)
= + 100 × 1.6 × 10^-19 × 14 = +2.24 10-16 J

C.

400 R

W_AB = nR
(T_f–T_i) = × − 2 R(500 300)

A.

B.

90 J

A.

6 cal

B.

diatomic

146 = Cv∆T
⇒ C_v = 21 J/mol K

B.

It introduces the concept of entropy

C.

1/3

A.

∆U₁ = ∆U₂

Internal energy is state function

B.

Internal energy and entropy are state functions.

In thermodynamic system, entropy and internal energy are state functions.

C.

There will be no change in number of moles if the vessels are joined by the valve. Therefore, from gas equation

Now, according to Boyle's law (pressure = constant) P₁ V₁ + P₂ V₂ = P(V₁ + V₂ )

D.

(a) 189 K (b) – 2.7 kJ

In an adiabatic Process

$$\begin{gathered} T{V}^{\gamma -1} = Cons\tan t \\ Or \\ {T}_1{V_1}^{\gamma -1} = T_2{V_2}^{\gamma -1} \\ For monoatomic gas \\ \gamma = \frac{5}{3} \\ \left(300\right)V^{2/3} = T_2(2V{)}^{2/3} \\ \Rightarrow T_2 = \frac{300}{(2{)}^{2/3}} \\ {T}_2 = 189 K\hspace{0.17em}(final temperature) \\ Change in internal energy, \\ ∆U = n\frac{f}{2}R∆T \\ = 2\left(\frac{3}{2}\right)\left(\frac{25}{3}\right)(-111) = -2.7 kJ \end{gathered}$$

A.

H = H₁+ H₂

B.

P → c, Q → a, R → d, S → b

Process I = Isochoric
II = Adiabatic
III = Isothermal
IV = Isobaric

B.

90 J

D.

r⁵

Power = rate of production of heat= F.V

$$\begin{gathered} = 6\pi \eta r{v}_{\tau }.V_{\tau } = 6\pi \eta r{V}_{\tau }^2 \\ (\because F = 6\pi \eta V_{\tau }r stoke\text{'}s formula) \\ {V}_{\tau } \propto r^2 \\ \because V_{\tau } = \frac{2}{9}\frac{r^2(\rho -\sigma )}{\eta }g \\ \therefore Power \propto r^5 \end{gathered}$$

B.

208.7 J

Applying first law of thermodynamics equation.

$$\begin{gathered} ∆Q = ∆U + ∆W \\ 54 x 4.18 \\ = ∆U + 1.013 x {10}^5 (167.1 x {10}^{-6} - 0) \\ (\because ∆W = P∆V) \\ ∆U = 208.7 J \end{gathered}$$

A.

2/5

Gas is monatomic, so

$C_p = \frac{5}{2}R$

Given process is isobaric

dQ = nC_pdt

$$\begin{gathered} \Rightarrow dQ = n\left(\frac{5}{2}R\right)dT \\ dW = PdV = nRdT \\ \therefore Required ratio = \frac{dW}{dQ} = \frac{nRdT}{n\left(\frac{5}{2}R\right)dT} = \frac{2}{5} \end{gathered}$$

A.

26.8%

Efficiency of ideal heat engine,

$\eta = \left(1-\frac{T_2}{T_1}\right)$

Sink temperature, T₂ = 100^oC = 100 + 273 = 373 K
Source temperature, T₁ = 0^o C = 0 + 273 = 273 K

Percentage efficiency,

$$\begin{gathered} \%\eta = \left(1-\frac{T_2}{T_1}\right) x 100 \\ = \left(1-\frac{273}{373}\right) x 100 - \left(\frac{100}{373}\right) x 100 = 26.8\% \end{gathered}$$

B.

O₂ and He

The slope of the adiabatic process of p-V diagram

$\left(\frac{\mathrm{dp}}{\mathrm{dv}}\right) = \mathrm{\gamma }\frac{\mathrm{p}}{\mathrm{V}} \mathrm{i}.\mathrm{e} \mathrm{slope} \mathrm{\alpha \gamma }$

From graph (slope)₂ > (slope)₁

⇒ γ₂ >γ₁

γ for monoatomic gas (Ar, He) is greater than γ for the diatomic gas ₍O₂, N₂)

A.

It is impossible

Efficiency of a heat engine is

$$\begin{gathered} \mathrm{\eta } = 1 - \frac{{\mathrm{T}}_1}{{\mathrm{T}}_2} \\ \mathrm{or} \\ \mathrm{\eta } = \frac{{\mathrm{T}}_1 -{\mathrm{T}}_2}{{\mathrm{T}}_1} \\ \mathrm{Here}, {\mathrm{T}}_1 = 273 + 127 = 400 \mathrm{K} \\ {\mathrm{T}}_2 = 273 + 27 = 300 \mathrm{K} \\ \therefore \mathrm{\eta } = \frac{400 -300}{400} = \frac{100}{400} = 0.25 \\ = 25\% \end{gathered}$$

Heat, 26% efficiency is impossible for a given heat engine.

B.

9.3×10^-11 %

Heat taken by the given body = mc Δθ ( specific heat of body =0.2 kcal/kg^oC ,  final temperature of the body =100^oC)

ΔE = m × 0.2 × 100 =100 m kcal

      =20m × 4.2 × 10³ J

Now gain of the mass is given by

$$\begin{gathered} ∆\mathrm{m} = \frac{∆\mathrm{E}}{{\mathrm{C}}^2} \left(\because \mathrm{E} ={\mathrm{mc}}^2\right) \\ =\frac{20\mathrm{m} \times 4.2 \times {10}^3}{{\left(3\times {10}^8\right)}^2} \\ \mathrm{Therefore} \mathrm{percentage} \mathrm{increase} \mathrm{in} \mathrm{mass} \mathrm{is} \mathrm{given} \mathrm{by} \\ \Rightarrow =\frac{∆\mathrm{m}}{\mathrm{m}}\times 100 \\ =\frac{20\mathrm{m} \times 4.2 \times {10}^3}{{\left(3\times {10}^8\right)}^2}\times 100 \\ =9.3\times {10}^{-11}\% \end{gathered}$$

A.

80 kg

Volume of the raft is given by

$$\begin{gathered} \mathrm{V} =\frac{\mathrm{m}}{\mathrm{d}}= \frac{120}{600} =0.2{\mathrm{m}}^3 \\ \mathrm{If} \mathrm{the} \mathrm{raft} \mathrm{is} \mathrm{immersed} \mathrm{fully} \mathrm{in} \mathrm{water}. \\ \mathrm{Then} \mathrm{weight} \mathrm{of} \mathrm{water} \mathrm{displaced} \mathrm{is} \\ = \mathbf{V}\mathbf{ }{\mathbf{d}}_{\mathbf{water}} \\ =0.2\times {10}^3 =200 \mathrm{kg} \\ \mathrm{Hence} \mathrm{the} \mathrm{additional} \mathrm{weight} \mathrm{which} \mathrm{is} \mathrm{supported} \mathrm{by} \mathrm{raft} : \\ 200 \mathrm{kg} - 120 \mathrm{kg} =\mathbf{ }\mathbf{80}\mathbf{ }\mathbf{kg} \end{gathered}$$

If the weight of the object is more than this it will sink.

A.

$\frac{2}{3}\mathrm{Q}$

The general principle of conservation of energy then implies that

$\mathbf{∆}\mathbf{Q}\mathbf{ }\mathbf{=}\mathbf{∆}\mathbf{W}\mathbf{ }\mathbf{+}\mathbf{∆}\mathbf{U}$    ......(1)

i.e the energy (ΔQ) supplied to the system to increase the internal energy of the system(ΔU) and rest in work on the environment (ΔW). Equation (1) known as First law of thermodynamics.

For monoatomic gas at constant pressure

$$\begin{gathered} \frac{∆\mathrm{U}}{\mathrm{Q}} =\frac{1}{3} \\ \mathrm{or} ∆\mathrm{U} =\frac{\mathrm{Q}}{3} \\ \mathrm{Now} \mathrm{applying} \mathrm{first} \mathrm{law} \mathrm{of} \mathrm{thermodyanamics} \\ \mathrm{W} =∆\mathrm{Q} -∆\mathrm{U} \\ =\mathrm{Q} -\frac{\mathrm{Q}}{3} \\ \mathrm{W} = \frac{2\mathrm{Q}}{3} \end{gathered}$$

C.

(2)^1/2 T

In adiabatic process the system is insulated from the surroundings and heat is absorbed or released. Adiabatic proceess of the ideal gas is PV^γ = constant

where γ is the ratio of specific heats at constant pressure and constant volume.

Gas obeys the equation

     PV^3/2 = constant

$$\begin{gathered} \mathrm{Comparing} \mathrm{with} {\mathrm{PV}}^{\mathrm{\gamma }}= \mathrm{constant} \\ \mathrm{\gamma } = \frac{3}{2} \end{gathered}$$

C.

No heat enters or leaves the system

In an isothermal process ( Temperature fixed), the ideal gas equation gives

PV = constant

i.e pressure of a given mass of gas varies inversely as its volume. In the isothermal process, heat enters or leaves the system, so keep the temperature constant, so statement (C) is wrong.

A.

Transfer of heat by radiation

When two bodies are at different temperatures and seperated by distance, the heat transfer between them is called as radiation heat transfer. A reversible process in which system and surroundings can be restored to the initial state from the final state without producing any changes in the thermodynamic properties called revrsible process.

(a) It is the process by which heat is transmitted from one place to another without heating the intervening medium. Hence it is not reversible process.

b) Nichrome wire is made of alloys and has high resistance. When current is passed through it, heat is produced. So, here electrical energy is converted into heat energy. Hence, it is not a
reversible process.

(c) It is the process by which heat is transmitted from one point to another through a substance in the direction of fall of temperature without the actual movement of the particles of the substances themselves. Hence, it cannot be reversible.

d) Isothermal compression is reversible, for example-Carnot cycle, Heat engine. Thus, choice (d) is correct.

C.

1.2  × 10⁴

The heat converted to work is the amount of heat that remains after going through the sink.
From the relation

$$\begin{gathered} \frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2} = \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} \\ \mathrm{Given} :- {\mathrm{Q}}_1 = 6 \times {10}^4 \mathrm{cal} \\ {\mathrm{T}}_1 = 227+ 273 = 500 \mathrm{K} \\ {\mathrm{T}}_2 = 127 + 273 = 400 \mathrm{K} \\ \therefore \frac{{\mathrm{Q}}_2}{6 \times {10}^4} = \frac{400}{500} \\ \Rightarrow {\mathrm{Q}}_2 = \frac{4}{5} \times 6 \times {10}^4 \mathrm{cal} \end{gathered}$$

Now heat converted to work:-

$$\begin{gathered} {\mathrm{Q}}_1 - {\mathrm{Q}}_2 \\ = 6.0 \times {10}^4 - 4.8 \times {10}^4 \\ = 1.2 \times {10}^4 \mathrm{cal} \end{gathered}$$

C.

Work

Work does not characterise the thermodynamic state of matter, it is a path function giving only relationship between two quantities.

A.

diatomic

In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero

For adiabatic process

       dQ = 0 

       dU = -ΔW

⇒ nC_V dT = + 146 × 10³ J

      [ f → degree of freedom ]

$\Rightarrow \frac{{10}^3 \times \mathrm{f} \times \times 8.3 \times 7}{2} = 146 \times {10}^3$

               f = 5.02 ≈ 5

So, it is diatomic gas.

A.

ΔU = -W in an adiabatic process

An isothermal process is a constant temperature process. In this process

T = constant (or) ΔT = 0

∴ΔU = nC_v  ΔT = 0

An adiabatic process is defined as one with no heat transfer into or out of a system. Therefore,

            Q = 0

From first law of thermodynamics

            W = - ΔU

            ΔU = -W

C.

decreases

The entropy function gives us a numerical measure of the irreversibility of a given process, ie, it is a measure of disorder of a system. During formation of ice cubes orderedness increases, ie, disorderness decreases, hence entropy decreases.

A.

( T - 4 ) K

In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero.

             PV^γ  = constant

Where γ is the ratio of specific heats ( ordinary or molar ) at constant pressure and at constant volume.

Work done by the gas in adiabatic process,

          W = $\frac{\mathrm{\mu } \mathrm{R} \left({\mathrm{T}}_{\mathrm{i}} - {\mathrm{T}}_{\mathrm{f}}\right)}{\mathrm{\gamma } - 1}$

∴         6 R = $\frac{1 \times \mathrm{R} \left(\mathrm{T} - {\mathrm{T}}_2\right)}{\left({\displaystyle \frac{5}{3} - 1}\right)}$

⇒         6 = $\frac{\mathrm{T} - {\mathrm{T}}_2}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

⇒        T - T₂ = 4

⇒         T₂ = ( T - 4) K

D.

$\frac{6\mathrm{T}}{5}$

The efficiency of a heat engine is defined as the ratio of work done to the heat supplied, ie,

       $\mathrm{\eta } = \frac{\mathrm{work} \mathrm{done}}{\mathrm{Heat} \mathrm{input} }$

           = $\frac{\mathrm{W}}{\mathrm{Q}}$

⇒    $\mathrm{\eta } = 1 - \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$

where T₂ is temperature of sink, and T₁ is temperature of hot reservoir.

∴    $\frac{40}{100} = 1- \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$

⇒       $\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} = 0.6$

⇒        ${\mathrm{T}}_2 = 0.6 {\mathrm{T}}_1$

Again,   $\frac{50}{100} = 1 - \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1\text{'}}$

⇒           $\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1\text{'}} = 0.5$

⇒            $\frac{0.6 {\mathrm{T}}_1}{{\mathrm{T}}_1\text{'}} = 0.5$

∴              ${\mathrm{T}}_1\text{'} = \frac{0.6}{0.5} {\mathrm{T}}_1$

⇒                ${\mathrm{T}}_1\text{'} = \frac{6}{5}\mathrm{T}$

D.

$\frac{3}{2}$

Given, that

           $\mathrm{\rho } \propto {\mathrm{T}}^3$

⇒     $\mathrm{\rho } {\mathrm{T}}^{ -3} = \mathrm{constant}$         .....(i)

Also for adiabatic process,

       $\mathrm{\rho } \mathrm{V}{ }^{\mathrm{\gamma }} = \mathrm{Constant}$

But  pV = RT (ideal gas law)

 ⇒    V = $\frac{\mathrm{RT}}{\mathrm{\rho }}$

∴     $\mathrm{\rho }{\left(\frac{\mathrm{RT}}{\mathrm{\rho }}\right)}^{\mathrm{\gamma }} = \mathrm{constant}$  

      ${\mathrm{\rho }}^{1-\mathrm{\gamma }} {\mathrm{T}}^{\mathrm{\gamma }}$ = constant             ..... (ii)

⇒     $\mathrm{\rho } {\mathrm{T}}^{\raisebox{1ex}{$\mathrm{\gamma }$}\!\left/ \!\raisebox{-1ex}{$\left(1-\mathrm{\gamma }\right)$}\right.}$ = constant 

Equating Eqs. (i) and (ii), we get

    $\frac{\mathrm{\gamma }}{1 - \mathrm{\gamma }} = -3$

⇒     $\mathrm{\gamma } = -3 + 3\mathrm{\gamma }$

⇒    2γ = 3

⇒     $\mathrm{\gamma } = \frac{3}{2}$

⇒     $\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}} = \frac{3}{2}$

B.

700 J

   For a diatomic gas

  C_V = $\frac{5}{2} \mathrm{R}$    and  C_P = $\frac{7}{2} \mathrm{R}$

The work done in an isobaric process is

    W = P ( V₂ - V₁ )

          = n R ( T₂ - T₁ )

⇒  T₂ - T₁ = $\frac{\mathrm{W}}{\mathrm{nR}}$

The heat given in an isobaric process is

      Q = n C_P (T₂ - T₁ )

      $\mathrm{n} {\mathrm{C}}_{\mathrm{P}} \frac{\mathrm{W}}{\mathrm{nR}} = \frac{7}{2} \mathrm{W}$  

⇒                 = $\frac{7}{2} \times 200 \mathrm{J}$

⇒     n C_P$\frac{\mathrm{W}}{\mathrm{nR}}$  = 700 J 

C.

2.08 × 10⁶ J

Given:-

Latent heat of vaporization = 2.25 × 10⁶ J kg^-1

P = 1atm = 100kPa = 1 × 10⁵ Pa 

Mass of the water = 1 kg

The volume of 1 kg water

⇒          V = $\frac{1}{1000} {\mathrm{m}}^3$

and volume of 1 kg steam 

⇒         V' = $\frac{1}{0.6} {\mathrm{m}}^3$

The increase in volume 

⇒           ΔV = ( V' - V )

⇒          ΔV  = $\frac{1}{0.6}$ m³$- \frac{1}{1000}$ m³

                   = ( 1.7 - 0.001 ) m³

⇒           ΔV = 1.699

⇒           ΔV ≈ 1.7 m³

The work done by steam is

      ΔW = P ΔV

      ΔW = ( 100 kPa ) (1.7 m³ )

      ΔW = 1.7 × 10⁵ J

The change in internal energy

      ΔU = ΔQ - ΔW

           = 2.25 × 10⁶ J - 1.7 × 10⁵ J

      ΔU = 2.08 × 10⁶ J  

D.

4.94 × 10³ J kg^-1

Given:-

          T_m = 130^o C

            h = 6.2 km = 6200 m

         S = 126 J kg^-1 oC^-1

Let the mass of the lead ball be m and its latent heat of fusion be L.

The initial temperature of the lead ball T₁ = 30^o C

The potential energy of the ball gets converted into heat melts the ball.

 ∴                   mgh = mS ( T_m - T₁ ) + mL

⇒   m (10) ( 6200 ) = m (126 ) (130 - 30 ) + mL

⇒                  6200 = 12600 + L

⇒                     L = 4.94 × 10⁴ J Kg^-1

A.

He and O₂

We know that

          $\left(\frac{\mathrm{dp}}{\mathrm{dV}}\right) = - \mathrm{\gamma } \left(\frac{\mathrm{p}}{\mathrm{V}}\right)$

∴          slope  ∝ γ

From the Figure, slope of 2 > slope of 1

So, plots 1 and 2 should correspond to He and O₂ respectively.

B.

40%

As we know that according to adiabatic expansion

          pV^γ = K

   γ is the ratio of specific heat.

By differentiating

∴        Δp V^γ + pγ V^γ-1 ΔV = 0

⇒          $\left(\frac{∆\mathrm{p}}{\mathrm{p}}\right) = - \mathrm{\gamma } \left(\frac{∆\mathrm{V}}{\mathrm{V}}\right)$
For monoatomc gas,

              $\mathrm{\gamma } = \frac{5}{3}$

            $\frac{∆\mathrm{p}}{\mathrm{p}} = \frac{5}{3} \times 24\hspace{0.17em}\%$

              $\frac{∆\mathrm{p}}{\mathrm{p}}$= 40 %

B.

$\frac{13 \mathrm{R}}{6}$

A pressure-volume ( or PV diagram or volume-pressure loop) is used to describe corresponding changes in volume and pressure in a system. They are commonly used in thermodynamics, cardiovascular physiology, and respiratory physiology.

${\mathrm{W}}_{\mathrm{A}\to \mathrm{B}} =$Area under p-V diagram

          = 4 V₀ × 3${\mathrm{p}}_0 + \frac{1}{2} 4 {\mathrm{V}}_0 \times 3 {\mathrm{p}}_0$

         = 18 p₀ V₀

∴  ΔU = nC_v ΔT

         = 1. $\frac{3}{2} \mathrm{R} \left({\mathrm{T}}_{\mathrm{B}} -{\mathrm{T}}_{\mathrm{A}}\right)$

         = $\frac{3}{2}\mathrm{R} \left(\frac{30 {\mathrm{p}}_0 {\mathrm{V}}_0}{\mathrm{R}} - \frac{3 {\mathrm{p}}_0 {\mathrm{V}}_0}{\mathrm{R}}\right)$

         = $\frac{81}{2 }{\mathrm{p}}_0 {\mathrm{V}}_0$

Thus,  $∆{\mathrm{Q}}_{\mathrm{A}\to \mathrm{B}} = ∆{\mathrm{W}}_{\mathrm{A}\to \mathrm{B}} + ∆{\mathrm{U}}_{\mathrm{A}\to \mathrm{B}}$

                       = 18 ${\mathrm{p}}_0 {\mathrm{V}}_0 + \frac{81}{2} {\mathrm{p}}_0 {\mathrm{V}}_0$

            $∆{\mathrm{Q}}_{\mathrm{A}\to \mathrm{B}}$ = $\frac{117}{2}{\mathrm{p}}_0 {\mathrm{V}}_0$

Molar heat capacity

            C = $\frac{∆\mathrm{Q}}{\mathrm{n} ∆\mathrm{T}}$

                = $\frac{117 {\mathrm{p}}_0 {\displaystyle \raisebox{1ex}{${\mathrm{V}}_0$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \frac{30 {\mathrm{p}}_0 {\mathrm{V}}_0}{\mathrm{R}} - \frac{3 {\mathrm{p}}_0 {\mathrm{V}}_0}{\mathrm{R}}}}$

               = $\frac{117 {\mathrm{p}}_0{\displaystyle \raisebox{1ex}{${\mathrm{V}}_0$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \frac{27 {\mathrm{p}}_0 {\mathrm{V}}_0}{\mathrm{R}}}}$

        C = $\frac{13\mathrm{R}}{6}$

B.

≈ 490 m/s

Given:-

 Specific heat S = 125 J/kg - K

 θ_i = 25^o C

 θ_m = 300^o C

 m = 2.5 × 10⁴ J/kg

Then,

 Heat energy supplied to lead bullet

  Q = 50% of KE

      = $\frac{1}{2} \left(\frac{1}{2}\mathrm{m} {\mathrm{v}}^2\right)$

  Q = $\frac{1}{4} {\mathrm{mv}}^2$

Heat energy utilised for rise of temp and melting of lead bullet

  Q' = mc Δθ+ mL

       = m ( c Δθ + L )

By equating Q and Q'

  $\frac{1}{4} {\mathrm{mv}}^2 = \mathrm{m} \left( \mathrm{c}∆\mathrm{\theta } + \mathrm{L}\right)$

⇒  $\frac{{\mathrm{v}}^2}{4} = 125 \left( 300 - 25 \right) + 2.5 \times {10}^4$  

⇒   $\frac{{\mathrm{v}}^2}{4}$= 59375 × 4

⇒   v² = $\sqrt{237500}$

⇒  v = 487.5 m/s

⇒  v = 490 m/s

C.

adiabatic curve slope = γ× isothermal curve slope

 For isothermal process pV = constant

⇒       $\left(\frac{\mathrm{dp}}{\mathrm{dV}}\right) = -\frac{\mathrm{p}}{\mathrm{V}}$ = slope of isothermal curve

For adiabatic process, pV^γ = constant

⇒        $\frac{\mathrm{dp}}{\mathrm{dV}} = -\frac{\mathrm{\gamma } \mathrm{p}}{\mathrm{V}}$ = slope of adiabatic curve

Clearly

           ${\left(\frac{\mathrm{dp}}{\mathrm{dV}}\right)}_{\mathrm{adiabatic}} = \mathrm{\gamma } {\left(\frac{\mathrm{dp}}{\mathrm{dV}}\right)}_{\mathrm{isothermal}}$

A.

Work done by the gas is positive

For isothermal process

   dU = 0

Work done = dW

                 = ρ ( v₂ $-$ v₁ )

∴  dW = + pV                          [ $\because$V₂ = 2V₁ = 2V ]

B.

3 / 5

By first law of thermodynamics

    ΔQ = ΔU + ΔW

Where ΔQ = Heat supplied to the system by the surroundings

          ΔW = Work done by the system on the surroundings

          ΔU = change in internal energy of the system

         = $\frac{\mathrm{f}}{2}$nRT + P $\int \mathrm{dV}$

   Q = $\frac{\mathrm{f}}{2}$ nRT + PV

   Q = $\frac{\mathrm{f}}{2}$ nRT + nRT                ( $\because$ PV = nRT )

      = $\frac{3}{2} \mathrm{nRT} + \mathrm{nRT}$

   Q = $\frac{5}{2}$ nRT

Fraction of heat energy supplied increase the internal energy.

∴  Fraction of heat energy supplied

                   = $\frac{\mathrm{U}}{\mathrm{Q}}$

                    = $\frac{{\displaystyle \left( \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \right) \mathrm{nRT}}}{{\displaystyle \left( \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \right) \mathrm{nRT}}}$

                     = $\frac{3}{5}$

B.

10 kJ

Efficiency, η = 1 $- \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$

              η = $1 - \frac{\left(273 + 27\right)}{\left(273 + 127\right)}$

              η  = 1 $- \frac{300}{400}$ 

                   = 1 $-\frac{3}{4}$

              η  = $\frac{1}{4}$

Also, The efficiency defination given by

        η  = $\frac{\mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{engine}}{\mathrm{Heat} \mathrm{supplied} \mathrm{by} \mathrm{source}}$

             = $\frac{\mathrm{W}}{40 \mathrm{kJ}}$

∴   W = 40 η 

           = 40 × $\frac{1}{4}$

     W = 10 kJ   

A.

If both assertion and reason are true and reason is the correct explanation of assertion.

In adiabatic process, no heat transfer takes place between gas and surrounding 

            ΔQ =  0

From definition of specific heat

            C = $\frac{∆\mathrm{Q}}{\mathrm{m} ∆\mathrm{T}}$ = 0

Again, for isothermal process

           ΔT =  0

∴          C = $\frac{∆\mathrm{Q}}{\mathrm{m} ∆\mathrm{T}}$ = ∞

B.

800 J

Total work done by gas from D to E to F is equal to the area of ΔDEF.

∴   Area of ΔDEF = $\frac{1}{2}$× DE × EF

Here   DF = change in pressure

               = 600 $-$ 200

               = 400 N m^-2

            EF = change in volume

                 = 7 m³ $-$ 3 m³

                  = 4 m³

Area of ΔDEF = $\frac{1}{2}$× 400 × 4

                    = 800 J

Thus, the total work done by the gas from D to E to F is 800 J.

C.

If assertion is true but reason is false.

An isolated system may have any process such as isothermal, adiabatic, cyclic etc.

C.

If assertion is true but reason is false.

In an adiabatic process, no exchange of heat is permissible

i.e                    dQ = 0.

As  dQ = dU + dW

           = 0

∴  dU = - dW.

Assertion is true but reason is false.