Thermodynamics

This is a long answer type question as classified in NCERT Exemplar

(a) For the process AB

dV=0 and dW=0

dQ=dU+dW=dU

dQ=dU= change in internal energy , so heat utilised is equal to change in internal energy.

Since p= $\frac{nR}{V}T,$ in adiabatic temperature is directly proportional to pressure. So heat is supplied to

This is a long answer type question as classified in NCERT Exemplar

pV^1/2= constant

P=k/ $\sqrt[]{V}$

Work done from 1 to 2

W= ${\int }_{V1}^{V2}pdV=K{\int }_{V1}^{V2}\frac{dV}{\sqrt[]{V}}=k{\left[\frac{\sqrt[]{V}}{\frac{1}{2}}\right]}_{V1}^{V2}=2K\left(\sqrt[]{V_2}-\sqrt[]{V_1}\right)$

from ideal equation = pV=nRT

T= pV/nR= $\frac{p\sqrt[]{V}\sqrt[]{V}}{nR}$

T= $\frac{K\sqrt[]{V}}{nR}$

T₁= $\frac{K\sqrt[]{V1}}{nR}$ , T₁= $\frac{K\sqrt[]{V2}}{nR}$

$\frac{T_1}{T_2}=\frac{\frac{k\sqrt[]{V_1}}{nR}}{\frac{k\sqrt[]{V_2}}{nR}}={\sqrt[]{\frac{V_1}{V_2}}}^{}$ = $\sqrt[]{\frac{V_1}{2V_1}}=\frac{1}{\sqrt[]{2}}$

U= $\frac{3}{2}RT$

$?U=U_2-U_1=\frac{3}{2}R\left(T_1-T_2\right)$

= $\frac{3}{2}$ RT₁( $\sqrt[]{V}-1$ )

$?W$ =2p₁V₁^1/2( $\sqrt[]{V_2}-\sqrt[]{V_2}$ )

 = 2p₁V₁^1/2(2 $\sqrt[]{V_1}-\sqrt[]{V_1}$ )

 = 2p₁V₁( $\sqrt[]{2}-1$ )= 2RT₁( $\sqrt[]{2}-1$ )

$?Q$ = $?U+?W$

= $\frac{3}{2}$ RT₁( $\sqrt[]{2}-1$ )+ 2RT₁( $\sqrt[]{2}-1$ )

= $\frac{7}{2}R{T}_1(\sqrt[]{2}-1)$

Temperature inside the refrigerator,  $T_1$ = 9 $?$ = 9 + 273 K = 282 K

Room temperature,  $T_2$ = 36 $?$ = 36 + 273 = 309 K

Coefficient of performance = $\frac{T_1}{\mathrm{}{T}_2-T_1}$ = $\frac{282}{309-282}$ = 10.44

Therefore, the coefficient of performance is 10.44

Total work done by the gas from D to E to F = Area of $?DEF$ = $\frac{1}{2}$ EF $\times DF$

Where DF = Change in pressure = 600 – 300 = 300 N/ $m^2$

FE = change in volume = 5-2 = 3 $m^3$

Area of $?DEF=$ $\frac{1}{2}\times$ 3 $\times 300$ = 450 J

Therefore work done by the gas from D to E to F is 450 J.

Heat is supplied to the system at a rate of 100W

Hence, heat supplied, Q = 100 J/s

The system performs at the rate of 75 J/s

Hence, work done, W = 75 J/s

From the 1st law of Thermodynamics, we have Q = U + W, where U is the internal energy

U = Q – W = 100 – 75 = 25 J/s = 25 W

Therefore the internal energy

Work done by the steam engine per minute, W = 5.4 $\times {10}^8$ J

Heat supplied by the boiler, H = 3.6 $\times {10}^9$ J

Efficiency of the engine,  $\eta$ = $\frac{Outputenergy}{Inputenergy}$ = $\frac{5.4\mathrm{}\times {10}^8}{3.6\mathrm{}\times {10}^9}$ = 0.15

Amount of heat wasted = Input energy – Output energy

= 3.6 $\times {10}^9-$ 5.4 $\times {10}^8=3.06\times {10}^9$ J

(a) When the stopcock is opened, the volume became double between cylinders A and B. Since volume is inversely proportional to pressure, the pressure will become half. So the initial pressure of 1 atm in cylinder A will become ½ atm in cylinder A and B.

(b) The internal energy will change whe

The work done, W = 22.3 J

Being an adiabatic process,  $?$ Q = 0

$?$ W = -22.3 J : since the work is done on the system

From the 1st law of thermodynamics, we know $?$ Q = $?U+$ $?$ W, where $?U$ is the change of internal energy of the gas

$?$ U = 22.3 J

When the gas goes f

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus the process is ‘Adiabatic’.

Let the initial and final pressure inside the cylinder be $P_1$ & $P_2$ and volume be $V_1$ & 

(a) When the two bodies at different temperatures brought in contact, heat flows from the body of higher temperature to the body with lower temperature till the thermal equilibrium is achieved and both the body attains the temperature of (T1 + T2 )/2. But only when the thermal capacities of both th