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CLASS NOTE OF THERMODYNAMIC WITH MORE PROBLEMS(QUESTIONS)

1. A certain gas at atmospheric pressure is compressed adiabatically so that its volume becomes half of its original volume. Calculate the resulting pressure in Nm^-2. Given γ for air = 1.4.

P₁ = 1 atm = 1.013 × 10⁵ Nm^-2, V₁/V₂ = 2

= 1.013 × 10⁵ × 2.64 = 2.694 × 10⁵ Nm^-2

2. A tyre pumped to a pressure of 6 atmosphere, suddenly bursts. Room temperature is 15^oC. Calculate the temperature of escaping air. γ = 1.4.

Here P₁ = 6 atm, T₁ = 15 + 273 = 288 K, γ = 1.4

P₂ = 1 atm

Taking log both sides, we get
1.4 log T₂ = 1.4 log 298 - 0.4 log 6

= 1.4(2.4742) - 0.4 (0.7782) = 3.1526

or

= 178.6 - 273 = 94.4^oC.

3. A quantity of air at 27^oC and atmospheric pressure is suddenly compressed to half its original volume. Find the final (i) pressure and (ii) temperature. Given γ for air = 1.42.

Here T₁ = 27 + 273 = 300 K, P₁ = 1 atm, V₂ = V₁/2,

γ = 1.42

(i) As P₂V₂^γ = P₁V₁^γ

or T₂ = 401.3 - 273 = 128.3^oC.

4. Dry air at 10 atm pressure and 15^oC is suddenly compressed to atmospheric pressure. Find the new temperature. Given γ = 1.41, log 2.88 = 0.4594, log 1.474 = 0.1686.

Here T₁ = 15 + 273 = 288 K, P₁/P₂ = 10

For adiabatic process,

Taking log of both sides

1.41 (log 288 - log T₂) = 0.41 log 10 = 0.41 × 1

T₂ = Antilog (2.1701) = 147.9 K or -125^oC

5. A Carnot engine operates between 227^oC and 127^oC. If it absorbs 60 × 10⁴ calorie at higher temperature, how much work per cycle can the engine perform?

Here T₁ = 227 + 273 = 500 K, T₂ = 127 + 273 = 400K

As

or W = 12 × 10⁴ cal

= 12 × 10⁴ × 4.2 J = 5.04 × 10⁵ J.

6.How much energy in watt hour may be required to convert 2 kg of water into ice at 0^oC, assuming that the refrigeration is ideal? Given temperature of freezer is -15^oC, room temperature is 25^oC and initial temperature of water is 25^oC.

Here T₁ = 25 + 273 = 298 K,

T₂ = -15 + 273 = 258 K

Specific heat of water, c = 4.2 × 10³ J kg^-1

Latent heat of ice, L = 3.36 × 10⁵ J kg^-1

Amount of heat required to be removed from 2 kg of water at 25^oC to change it into ice at 0^oC,

Q₂ = Mc(θ₂ - θ₁) + ML

= 2 × 4.2 × 10³ (25 - 0) + 2 × 3.36 × 10⁵

= 2.1 × 10⁵ + 6.72 × 10⁵

= 8.82 × 10⁵J

Heat rejected to the surroundings,

Energy supplied to convert water into ice,

W = Q₁ - Q₂ = (10.15 - 8.82) × 10⁵ = 1.33 × 10⁵ J

 =

7. Assuming that a domestic refrigerator can be regarded as a reversible engine working between the temperature of melting ice and that of the atmosphere (17^oC), calculate the energy which must be supplied to freeze one kilogram of water already at 0^oC.

Here T₂ = 273 K, T₁ = 17 + 273 = 290 K

Heat required to be removed to freeze 1 kg of water at 0^oC

Q₂ = mL = 1 × 80,000 cal = 80,000 × 4.2 J

8.A refrigerator whose coefficient of performance is 5 extracts heat from the cooling compartment at the rate of 250 J/cycle?How much electric energy is spent per cycle? How much heat per cycle is discharged to the room?

Here Q₂ = 250J/cycle, β = 5

Heat discharged to the room per cycle,

Q₁ = Q₂ + W = 250 + 50 = 300 J.

9. A refigerator has to transfer an average of 263 J of heat per second from temperature -10^oC to 25^oC. Calculate the average power consumed, assuming no energy losses in the process.

Here T₁ = 25 + 273 = 298 K,

T₂ = 10 + 273 = 263 K,

Q₂ = 263 J

Average power consumed,

W = Q₁ - Q₂ = 298 - 263 = 35 Js^-1 = 35 W.

10.An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q₁ = 5960 J, Q₂ = -5585J, Q₃ = -2980 J and Q₄ = 3645 J respectively. The corresponding works involved are W₁ = 2200 J, W₂ = - 825 J, W₃ = -1100 J and W4 respectively. Find (i) W₄ and (ii) efficiency of the cycle.

(i) By the first law of thermodynamics,

dQ = dU + dW

But for a cyclic process, dU = 0

∴ dQ = dW

or Q₁ + Q₂ + Q₃ + Q₄ = W₁ + W₂ + W₃ + W₄

or 5960 - 5585 - 2980 + 3645 = 2200 - 825 - 1100 + W₄

or W₄ = (5960 + 3645 + 825 + 1100) - (5585 + 2980 + 2200)

= 11530 - 10765 = 765 J.

(ii) Efficiency,

11. Let the temperatures T1 and T2 of the two heat reservoirs in an ideal carnot engine be 1500^oC and 500^oC respectively. Which of these, increasing T₁ by 100^oC or decreasing T₂ by 100^oC, would result in a greater improvement in the efficiency of the engine?

Efficiency of Carnot engine,

(i) When T₁ is increased form 1500^oC to 1600^oC or 1600 + 273 = 1873 K and T₂ remains constant i.e., 500^oC or 500 + 273 = 773 K, we have

(ii) When T₁ remains constant i.e., 1500^oC or 1500 + 273 = 1773 K and T₂ is decreased by 100^oC from 500^oC to 400^o or 400 + 273 = 673 K, we have

Thus, η₂ > η₁.

Hence efficiency will increase if T₂ is decreased from 500^oC to 400^oC.