Questions: Thermodynamics

Q1. State the first law of thermodynamics and explain the significance of each term.

The first law states that the heat supplied to a system equals the increase in its internal energy plus the work done by the system: dQ = dU + dW. The term dQ is the energy entering the system as heat, taken positive when the system absorbs heat. The term dU is the change in internal energy, a quantity fixed by the initial and final states. The term dW is the work done by the system, taken positive when the system expands against its surroundings. The law is the energy balance extended to include heat, and it can be written with specific heats for a gas. In a cycle dU returns to zero, so the net heat supplied equals the net work done.

Hinglish: First law energy conservation ka extended form hai: dQ = dU + dW. Heat system me aati hai, kuch internal energy badhane me jaati hai aur kuch work karne me. Cycle me dU zero hota hai, isliye net heat net work ke barabar hoti hai.

Q2. Why is C_P greater than C_V for an ideal gas?

At constant volume the gas does no work, so all the heat supplied goes into raising the internal energy and the temperature. At constant pressure the gas expands as it is heated, so the same temperature rise needs extra heat to do the expansion work P delta V. Therefore C_P exceeds C_V by the work needed for the same temperature rise, and for one mole of an ideal gas the difference is exactly R, the gas constant. The ratio gamma = C_P/C_V is therefore greater than one, with values of about 1.67 for a monatomic gas and 1.4 for a diatomic gas.

Hinglish: Constant volume par work zero hota hai, isliye saari heat internal energy me jaati hai. Constant pressure par gas expand hota hai aur extra work karta hai, isliye Cp bada hota hai. Difference R ke barabar hai, aur gamma = Cp/Cv hamesha 1 se bada.

Q3. Show that the work done by an ideal gas at constant temperature is nRT ln(V2/V1).

At constant temperature the product PV equals nRT, so the product is fixed along the path. For a small change of volume dV the gas does P dV of work, and with P = nRT/V this becomes dW = nRT dV/V. Adding these contributions from V1 to V2 gives W = nRT ln(V2/V1). The value is positive for an expansion and negative for a compression. For an ideal gas the internal energy is a function of temperature alone, so delta U = 0 here and the heat taken in equals the work done.

Hinglish: Constant temperature par PV = nRT fix rehta hai, isliye P = nRT/V. Work integral nRT ln(V2/V1) deta hai. Ideal gas ki internal energy sirf temperature par depend karti hai, isliye delta U zero hota hai aur Q = W ho jaata hai.

Q4. A gas expands from 2 litres to 5 litres at a constant pressure of 1.5 x 10^5 Pa. Find the work done and the heat supplied if the internal energy rises by 300 J.

The work done at constant pressure is P delta V = 1.5 x 10^5 x (5 - 2) x 10^-3 = 1.5 x 10^5 x 3 x 10^-3 = 450 J. From the first law the heat supplied is dQ = dU + dW = 300 + 450 = 750 J. The gas converts part of the heat into expansion work and the rest into internal energy. If the volume had been kept constant, the same 300 J rise in internal energy would need only 300 J of heat.

Hinglish: Constant pressure par work P delta V se nikalta hai, yahan 450 J. First law se heat dU + dW = 300 + 450 = 750 J. Volume constant hota to sirf 300 J heat lagti.

Q5. What is an adiabatic process? Show that an adiabatic curve is steeper than an isothermal curve through the same point.

An adiabatic process exchanges no heat with the surroundings, so dQ = 0 and the first law gives dU = -dW. For an ideal gas, PV^gamma is constant along an adiabatic path. Differentiating PV = constant for the isotherm gives dP/dV = -P/V. Differentiating PV^gamma = constant gives dP/dV = -gamma P/V. Since gamma is greater than one, the magnitude of the slope is larger for the adiabatic curve, so it falls faster than the isotherm through the same point. This is why a gas compressed adiabatically heats up more than one compressed isothermally.

Hinglish: Adiabatic me dQ zero hota hai, isliye dU = -dW. PV^gamma constant ko differentiate karo to slope -gamma P/V milta hai, jo isotherm -P/V se zyada steep hai. Isi liye adiabatic compression me gas zyada garam hoti hai.

Q6. State the second law of thermodynamics and give one example that shows it.

One form of the second law says heat does not flow by itself from a colder body to a hotter body. A second form says no cyclic engine can convert all the heat it absorbs into work; some heat must be rejected to a colder reservoir. Both statements forbid the same class of processes and are equivalent. A simple example is a cup of hot tea in a room: it cools by losing heat to the room, but the room never gives heat back to heat the tea again, although the first law would allow that. A refrigerator can move heat from cold to hot, but only because work is done on it by a compressor.

Hinglish: Second law kehta hai heat apne aap cold se hot body me nahi jaati, aur koi engine poori heat ko work me convert nahi kar sakta. Chai thandi hoti hai lekin room se heat lekar garam nahi hoti; refrigerator me heat ko ulta bhejne ke liye work karna padta hai.

Q7. A Carnot engine works between 500 K and 300 K. Find its efficiency and the heat rejected if it absorbs 2.5 kJ per cycle.

The efficiency is 1 - T2/T1 = 1 - 300/500 = 0.40, so 40 percent of the heat absorbed becomes work. With Q1 = 2.5 kJ, the work is 0.40 x 2.5 = 1.0 kJ, and the heat rejected is Q2 = Q1 - W = 1.5 kJ. The same result follows from the reversible heat ratio Q1/T1 = Q2/T2, which gives Q2 = 2.5 x 300/500 = 1.5 kJ.

Hinglish: Carnot efficiency 1 - T2/T1 se 0.4 aati hai. Q1 2.5 kJ hai to work 1.0 kJ aur rejected heat 1.5 kJ. Q1/T1 = Q2/T2 se bhi wahi answer nikalta hai.

Q8. Explain the coefficient of performance of a refrigerator and calculate it for a reversible refrigerator between 270 K and 300 K.

The coefficient of performance is the heat removed from the cold space divided by the work supplied, beta = Q2/W. For a reversible refrigerator between cold temperature T2 and hot temperature T1, it equals T2/(T1 - T2). With T2 = 270 K and T1 = 300 K, beta = 270/30 = 9. So nine units of heat are removed for every unit of work supplied by the motor. A real refrigerator has a smaller coefficient because of irreversible losses, and the value falls as the temperature difference grows.

Hinglish: Refrigerator ka COP Q2/W hota hai. Reversible ke liye beta = T2/(T1 - T2). Yahan 270/(300-270) = 9 aata hai. Real refrigerator ka COP isse kam hota hai kyunki losses hote hain.

Q9. Why does a gas cool on adiabatic expansion?

In an adiabatic expansion no heat enters the gas, but the gas does work against the surroundings. By the first law, dU = -dW, so the internal energy falls and the temperature falls with it. For an ideal gas the temperature change follows T1V1^(gamma-1) = T2V2^(gamma-1), so an increase in volume at constant entropy lowers the temperature. This is the principle behind the cooling of air in a bicycle pump when it is released quickly, and the reason why adiabatic demagnetisation can cool a sample to very low temperatures.

Hinglish: Adiabatic expansion me gas work karta hai lekin heat nahi aata, isliye internal energy girta hai aur temperature girti hai. T V^(gamma-1) constant se yeh change nikalta hai.

Q10. In a cyclic process a gas absorbs 600 J of heat, does 200 J of work and rejects the rest. Is the cycle possible, and what is its efficiency?

For a cycle the change in internal energy is zero, so the net heat absorbed equals the net work done. The gas absorbs 600 J, does 200 J of work and rejects 400 J to the cold reservoir. The efficiency is W/Q1 = 200/600 = 0.33, about 33 percent. A 100 percent efficiency would need the rejected heat to be zero, which the second law forbids for a cycle between two reservoirs. The cycle is possible, and its efficiency must also not exceed the Carnot value for the two temperatures involved.

Hinglish: Cycle me dU zero hota hai, isliye net heat = net work. Gas 600 J heat leta hai, 200 J work karta hai aur 400 J reject karta hai. Efficiency 200/600 = 33 percent. 100 percent efficiency second law ke against hai.

Quiz below: five questions on the first law, specific heats, processes, Carnot efficiency and the second law. Try them before checking the formula sheet.