Solution-Class-12-Physics-Chapter-4-Thermodynamics-MSBSHSE

(B) 4J

(B) While melting, an ice cube remains at the same temperature. [Here, ∆u = 0, W = Q]
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.

(A) T_H is large
(B) T_C is low
(C) T_H – T_C is large
[ η = \(\frac{T_H-T_C}{T_H}=1-\frac{T_C}{T_H}\), ]

(B) energy

closed tube [refer the textbook]

If a gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed then

(a) There is no transfer of heat.

(b) The work is done on the gas

Hot water in a container cools after sometime. Its temperature goes on decreasing with time and after sometime it attains room temperature.

Note ; Here, we do not provide heat to the water or remove heat from the water. The water cools on exchange of heat with the surroundings.

(a) Melting of ice (b) Boiling of water

The temperature of the cold reservoir sets the limit on the efficiency of a heat engine.

[Note: η = \(1-\frac{T_C}{T_H}\)  This formula shows that for maximum efficiency, T_C should be as low as possible and T_H should be as high as possible.]

For a Carnot engine, efficiency

η = \(1-\frac{T_C}{T_H}\),  η →1 as T_C → 0

With two isothermal and two adiabatic processes all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs.

Thus, the cycle includes two isothermal and two adiabatic processes for maximum efficiency.

The internal energy of a system is the sum of potential energy and kinetic energy of all the constituents of the system. In the example stated above, conversion of potential energy into kinetic energy is responsible for a considerable rise in pressure and temperature of the mixture of hydrogen and oxygen ignited by the spark.

(a) Heat is generated into the resistor due to the passage of electric current. In the usual notation, heat generated = I²Rt.

(b) Yes. Water receives heat from the resistor.

(c) I²Rt  = MCΔT + PΔV

(Q)      (ΔU)      (W)

Here, I = current through the resistor, R = resistance of the resistor, t = time for which the current is passed through the resistor, M=mass of the water, S =specific heat of water, T=rise in the temperature of water, P = pressure against which the work is done by the water, ΔU = increase in the volume of the water.

(a) Heat has been transferred from the chamber to the water bath.

(b) No work is done by the system (the mixture of fuel and oxygen) as there is no change in its volume.

(c) There is increase in the temperature of water. Therefore, ΔU is positive for water. For the system (the mixture of fuel and oxygen), ΔU is negative.

Consider some quantity of an ideal gas enclosed in a cylinder fitted with a movable, massless and frictionless piston. See Fig..

Suppose the gas is allowed to expand by moving the piston outward extremely slowly. There is decrease in pressure of the gas as the volume of the gas increases. Figure shows the corresponding p— V diagram.

In this case, the work done by the gas on its surroundings,

W =\( \int_{V_1}^{V_2}PdV\) (= area under the curve) is positive as the volume of the gas has increased from V₁ to V₂.

Let us now suppose that starting from the same initial condition, the piston is moved inward extremely slowly so that the gas is compressed.

There is increase in pressure of the gas as the volume of the gas decreases.

Below figure shows the corresponding p—V diagram.

In this case, the work done by the gas on its surroundings,

W = \( \int_{V_1}^{V_2}PdV\)  ( = area under the curve) is negative as the volume of the gas has decreased from V₁ to V₂ .

Similarities :

• Heat is added to the system.
• There is increase in the internal energy of the system.
• Work is done by the system on its environment.

Differences :

• In a solar cooker, heat is supplied in the form of solar radiation. The rate of supply of heat is relatively low.
• In a pressure cooker, usually LPG is used (burned) to provide heat. The rate of supply of heat is relatively high.
• As a result, it takes very long time for cooking when a solar cooker is used. With a pressure cooker, it does not take very long time for cooking.

Note ; A solar cooker can be used only when enough solar radiation is available

Data: P = 1 atm = 1.013 x 10⁵ Pa, V₁ =5 litres =5 x 10^-3 m³,

V₂ = 10 Litres = 10 x 10^-3 m³, Q = 400J.

The work done by the system (gas in this case) on its surroundings,

W = P (V₂ − V₁)

= (1.013 x 10⁵) (10 x 10^-3 − 5 x 10^-3)

= 1.013 (5 x 10²)] = 5.065 x 10²J

The change in the internal energy of the system,

ΔU = Q − W = 400 − 506.5 = −106.5J

The minus sign shows that there is a decrease in the internal energy of the system.

Given: Q = −125kJ, W= −104kJ

ΔU = Q − W= −125−(−104)

= (−125 + 104) = −21kJ. Negative sign shows that decrease in energy.

This is the change (decrease) in the internal energy.

Given : η = 75% = 0.75, T_H = (273 + 727) K = 1000 K

η = \(1−\frac{T_C}{T_H}\) ∴  \frac{T_C}{T_H}= 1− η

T_C = T_H (1 − η ) = 1000(1 - 0.75)

= 250 K = (250 − 273)°C

= −23°C

This is the temperature of the cold reservoir.

Given : T_C = 250 K, T_H = 300 K
K = \(\frac{T_C}{T_H-T_C}=\frac{250}{300-250}=\frac{250}{50}\)  = 5
this is the coefficient of performance of the refrigerator.

Given : W = 2000J , isothermal process

In this case, the change in the internal energy of the gas, ΔU=0,  as the gas is taken through an isothermal process.

Hence, the heat added to it,

Q = ΔU + W = 0 + W = 2000J

(Note : For an isothermal process, none of the quantities Q and W is zero.)

Given : T_f = 2T_i, monatomic gas  ϒ = 5/3

P_iV_i^ϒ = P_fV_f^ϒ in an adiabatic process

Now, PV= nRT  ∴ V = nRT/P

∴ V_i = \(\frac{nRT_i}{P_i}\) and V_f = \(\frac{nRT_f}{P_f}\)

∴ \(P_i(\frac{nRT_i}{P_i})^\gamma = P_f(\frac{nRT_f}{P_f})^\gamma\)

∴ P_i^1−ϒT_i^ϒ = P_f^1−ϒT_f^ϒ

∴ \((\frac{T_f}{T_i})^{\gamma}\) = \((\frac{P_i}{P_f})^{1-\gamma}\)

∴ \((\frac{T_f}{T_i})^{\gamma}\) = \((\frac{P_f}{P_i})^{\gamma -1}\)

∴ \(2^\frac{5}{3} = (\frac{P_f}{P_i})^{\frac{5}{3}-1} = (\frac{P_f}{P_i})^{\frac{2}{3}}\)

∴ \(\frac{5}{3}log\,2=\frac{2}{3}log(\frac{P_f}{p_i})\)

∴ \(\frac{5}{3}0.3010=\frac{2}{3}log(\frac{P_f}{p_i})\)

∴ \((2.5)(0.3010)=log(\frac{P_f}{p_i})\)

∴ 0.7525 = \(log(\frac{P_f}{p_i})\)

= \((\frac{P_f}{p_i})\) = antilog (0.7525) = 5.656

This is ratio of final pressure to its initial pressure

Given : a = \(\frac{ΔV_{max}}{2}=\frac{6-2}{2}×10^{-3}m^3\)  = 2 x 10⁻³ m³,

b = \(\frac{ΔP_{max}}{2}=\frac{11-1}{2}×10^{5}Pa\) = 5 x 10⁵ Pa

Cycles = 25

The work done in one cycle, \(\oint PdV\)

= πab = (3.142) (2 x 10⁻³) (5 x 10⁵)

= 3.142 x 10³J

Hence, the work done in 25 cycles

= (25) (3.142 x 10³) = 7.855 x 10⁴J

This is the work done in 25 cycles

(a) P—- V diagram (Schematic):

ab : isobaric process, bc : isothermal process, cd : isobaric process, da : isothermal process.

\(\frac{P_aV_a}{T_a}=\frac{P_bV_b}{T_b}=\frac{P_cV_c}{T_c}=\frac{P_dV_d}{T_d}\)= nR

(b) P—T diagram (Schematic) :

Given : P_A = P_B = 6 x 10⁵ Pa, P_C = P_D= 2 x 10⁵ Pa

V_A = V_D = 2 L, V_B = V_C = 6 L, 1 L=10⁻³ m³

(i) The work done along the path A —> B (isobaric process), W_AB = P_A (V_B − V_A) =

(6 x 10⁵) (6 − 2) (10^-3) = 2.4 x 10³J

(ii) W_BC = zero as the process is isochoric (V = constant).

(iii) The work done along the path C —> D (isobaric process), W_CD = P_C (V_D− Vc)

= (2 x 10⁵ Pa) (2 − 6) (10^-3) = −8 x 10²J

(iv) W_DA = zero as V = constant.