Questions: Thermodynamics

Q1 What is a state function? Give two examples and explain why heat is not one.

Answer (English): A state function is a property whose value depends only on the current state of the system and not on the path used to reach that state. Internal energy, enthalpy, entropy and Gibbs energy are state functions. Heat is not a state function because the amount of heat exchanged depends on how the change is carried out: heating a gas at constant volume or at constant pressure from the same initial state to the same final temperature gives the same final state but different amounts of heat. The difference between the heat exchanged in the two paths is exactly the work done by the gas in one of them.

Hinglish: State function sirf current state par depend karta hai, path par nahi; U, H, S, G examples hain. Heat state function nahi hai kyunki same initial aur final state tak pahunchne ke different paths mein heat exchange alag hota hai (constant V vs constant P); yeh farq work se aata hai.

Q2 State the first law of thermodynamics and apply it to a gas that absorbs 500 J of heat and does 200 J of work.

Answer (English): The first law says that the change in internal energy of a closed system equals the heat supplied to it plus the work done on it. In the chemistry convention heat absorbed is positive and work done on the system is positive. Here q = +500 J and the gas does 200 J of work, so w = -200 J. The change in internal energy is 500 - 200 = +300 J. The internal energy rises because the heat supplied is larger than the energy spent as work.

Hinglish: First law ΔU = q + w. Heat absorbed q = +500 J, aur gas ne 200 J work kiya to w = -200 J. ΔU = 500 - 200 = +300 J; heat work se zyada hai isliye internal energy badhi.

Q3 Calculate the work done when 2 mol of an ideal gas expands isothermally and reversibly from 10 L to 20 L at 300 K.

Answer (English): For an isothermal reversible expansion of an ideal gas, the work done on the gas is minus nRT ln(V2/V1). Substituting n = 2 mol, R = 8.314 J K^-1 mol^-1, T = 300 K and the volume ratio 2 gives w = -2 x 8.314 x 300 x ln 2. Since ln 2 is 0.693, the value is -3.46 x 10^3 J, so the gas does about 3.46 kJ of work on the surroundings. The negative sign follows the convention that work done by the system is negative.

Hinglish: w = -nRT ln(V2/V1) = -2 × 8.314 × 300 × ln 2 = -3.46 × 10^3 J. Negative sign matlab work system ne kiya, surroundings par. Reversible expansion mein work maximum hota hai.

Q4 Explain why delta H is usually equal to the heat exchanged in an open beaker but delta U in a sealed bomb.

Answer (English): In an open beaker the pressure stays at atmospheric value while the volume changes, so the heat exchanged at constant pressure equals the enthalpy change. In a sealed bomb the volume is fixed, so no pressure-volume work is possible and the heat exchanged equals the internal energy change. The two quantities differ by the pressure-volume term, and the difference is written as delta n_g RT for a reaction involving gases.

Hinglish: Open beaker mein pressure constant rehta hai, isliye heat = ΔH. Sealed bomb mein volume fixed hota hai, isliye work zero hota hai aur heat = ΔU. Farq pV term ka hai, gases ke liye Δn_g RT.

Q5 Use the given combustion data to find the enthalpy of formation of methane. The combustion enthalpies of methane, graphite and hydrogen are -890.3, -393.5 and -285.8 kJ mol^-1.

Answer (English): The formation equation is C(graphite) + 2H2(g) giving CH4(g). Use Hess's law with the combustion equations: methane combustion releases -890.3, carbon combustion -393.5, and the combustion of two moles of hydrogen gives 2 x (-285.8). The target formation enthalpy equals the sum of the combustion enthalpies of the elements minus the combustion enthalpy of methane: -393.5 + 2(-285.8) - (-890.3), which is -74.8 kJ mol^-1. The negative value shows that methane is more stable than its elements at standard conditions.

Hinglish: Formation equation C(graphite) + 2H2 → CH4. Hess's law se ΔfH = ΔcH(C) + 2ΔcH(H2) - ΔcH(CH4) = -393.5 - 571.6 + 890.3 = -74.8 kJ/mol. Negative matlab methane elements se zyada stable hai.

Q6 A reaction has delta H = -92 kJ mol^-1 and delta S = -198 J K^-1 mol^-1. At what temperature does it become spontaneous?

Answer (English): Use delta G = delta H - T delta S and set delta G equal to zero to find the balance temperature. Convert delta H to joules so the units match: -92000 J mol^-1. Then T = delta H / delta S = -92000 / -198, which is about 465 K. Because delta H is negative and delta S is negative, the enthalpy term dominates below 465 K and the reaction is spontaneous there. Above 465 K the entropy term dominates and the reaction becomes non-spontaneous.

Hinglish: T = ΔH/ΔS = -92000/-198 ≈ 465 K. ΔH aur ΔS dono negative hain, isliye 463 K se neeche reaction spontaneous hai; high temperature par entropy term dominate karti hai aur reaction non-spontaneous ho jaati hai.

Q7 Explain the significance of the second law of thermodynamics and the meaning of entropy.

Answer (English): The second law fixes the direction of spontaneous change: the total entropy of the universe increases. Entropy measures the dispersal of energy and the number of equivalent arrangements available to a system. A gas spreading through a container has many more arrangements than a gas confined to one half, so its entropy is higher. A solid melting or a liquid boiling increases entropy because the particles gain freedom. The law explains why heat flows from hot to cold and why a spontaneous process never reverses without work being done on the system.

Hinglish: Second law direction batata hai: universe ki entropy spontaneous change mein badhti hai. Entropy energy ke dispersal aur arrangements ki sankhya ka measure hai. Gas ka container mein failna, solid ka pighalna, liquid ka ubalna , sab entropy badhate hain. Isi law se heat ka hot se cold flow aur spontaneous process ki irreversibility samajh aati hai.

Q8 For the reaction N2(g) + 3H2(g) giving 2NH3(g), delta H is -92.4 kJ mol^-1 and delta S is -198 J K^-1 mol^-1. Calculate delta G at 298 K and state whether the reaction is spontaneous.

Answer (English): Use delta G = delta H - T delta S. The enthalpy term is -92.4 kJ mol^-1, which is -92400 J mol^-1. The entropy term is 298 x (-198) = -59004 J mol^-1, so delta G = -92400 - (-59004) = -33396 J mol^-1, about -33.4 kJ mol^-1. The value is negative, so the synthesis of ammonia is spontaneous at 298 K under standard conditions. The entropy change is negative because four moles of gas become two, but the large negative enthalpy term outweighs it at this temperature.

Hinglish: ΔG = ΔH - TΔS = -92400 - 298(-198) = -33.4 kJ/mol. Negative hai isliye 298 K par reaction spontaneous. Gases 4 mol se 2 mol ho rahe hain isliye ΔS negative hai, par ΔH ka negative term zyada strong hai.

Q9 What is the relation between the standard Gibbs energy change of a reaction and its equilibrium constant?

Answer (English): The relation is delta G standard equal to minus RT ln K. A large negative delta G standard makes ln K large and positive, so K is much greater than one and the products dominate at equilibrium. A large positive delta G standard makes K much smaller than one and the reactants dominate. When delta G standard is zero, K equals one and the products and reactants are present in comparable amounts at equilibrium. The relation connects the thermodynamics of the standard state with the equilibrium position measured in the laboratory.

Hinglish: ΔG° = -RT ln K. ΔG° bahut negative ho to K bahut bada hota hai aur products dominate karte hain; ΔG° positive ho to K chhota aur reactants dominate karte hain. ΔG° = 0 par K = 1 hota hai. Yeh relation thermodynamics ko equilibrium se jodta hai.

Q10 A gas expands from 1 L to 5 L against a constant external pressure of 2 bar. Calculate the work done in joules.

Answer (English): The work done on the gas is minus the external pressure times the change in volume: w = -p_ext (V2 - V1). The volume change is 4 L, and one litre bar equals 100 J, so w = -2 x 4 x 100 = -800 J. The negative sign shows that the gas does work on the surroundings. If the same expansion were carried out reversibly and isothermally, the magnitude of the work would be larger, because the gas would push against a slightly smaller external pressure throughout the path.

Hinglish: w = -p_ext ΔV = -2 bar × 4 L = -8 L bar = -800 J (1 L bar = 100 J). Negative sign matlab gas ne surroundings par work kiya. Reversible path mein work ka magnitude zyada hota hai.

That closes the chapter questions. Try the quick quiz below before moving on to the previous year questions.