First Law Thermodynamics

Answer: D. ΔU = q + w = 500 + 200 = 700 J. Heat absorbed is positive, work done on the system is positive in chemistry convention (NCERT Class 11 Chemistry Chapter 5, page 139).

Why A is wrong: A (300 J) results from subtracting w from q: ΔU = 500 − 200. This uses the physics convention (ΔU = q − w) instead of the chemistry convention (ΔU = q + w).

Why B is wrong: B (−700 J) results from making both q and w negative before adding. When the problem states heat is absorbed and work is done on the system, both are positive in chemistry convention.

Why C is wrong: C (−300 J) results from reversing both signs: ΔU = −q + w or similar misapplication. Neither sign should be flipped when both q and w are already stated with their chemistry-convention signs.

Answer: C. In chemistry convention, w is work done ON the system. The gas does 150 J of work on the surroundings, so the surroundings do −150 J on the system. Therefore w = −150 J (NCERT Class 11 Chemistry Chapter 5, page 139).

Why A is wrong: A (+150 J) confuses work done by the system with work done on the system. Chemistry convention defines w as work done on the system; expansion work by the gas is negative.

Why B is wrong: B (0 J) would apply only if there were no volume change (isochoric process). Here the gas explicitly expands, so w ≠ 0.

Why D is wrong: D (+300 J) has no basis — it doubles the magnitude and gets the sign wrong. No operation in ΔU = q + w produces this from 150 J of expansion work.

Answer: D. In a cyclic process the system returns to its initial state. Since internal energy is a state function, ΔU = 0. From ΔU = q + w: 0 = q + w, so q = −w (NCERT Class 11 Chemistry Chapter 5, page 139).

Why A is wrong: A (q = 0) would apply to an adiabatic process, not a cyclic one. A cyclic process can exchange heat — the constraint is ΔU = 0, not q = 0.

Why B is wrong: B (w = 0) would apply to an isochoric (constant volume) process. In a cyclic process, work can be non-zero as long as q = −w.

Why C is wrong: C (ΔU = q) applies when w = 0 (isochoric), not in a cyclic process. In a cycle, ΔU = 0 regardless of q and w individually.

Answer: B. Internal energy depends only on the state of the system, not on how the state was reached. Heat and work are path functions — their values depend on the process path (NCERT Class 11 Chemistry Chapter 5, page 139).

Why A is wrong: A is incorrect. Heat (q) is a path function — the amount of heat exchanged depends on the process (adiabatic vs isothermal paths between the same states give different q).

Why C is wrong: C is incorrect. Work (w) is a path function — reversible and irreversible expansion between the same states produce different amounts of work.

Why D is wrong: D is incorrect. Neither q nor w is a state function. Only ΔU (the combination q + w) is path-independent.

Answer: A. Adiabatic means q = 0. So ΔU = q + w = 0 + 300 = +300 J. Work done on the system is positive in chemistry convention (NCERT Class 11 Chemistry Chapter 5, page 139).

Why B is wrong: B (0 J) would be correct for a cyclic process (ΔU = 0), not an adiabatic one. Adiabatic sets q = 0, not ΔU = 0.

Why C is wrong: C (−300 J) reverses the sign of w. When work is done ON the gas (compression), w = +300 J in chemistry convention, not −300 J.

Why D is wrong: D (+600 J) doubles the value with no justification. In ΔU = q + w with q = 0, there is no factor of 2.

Answer: A. At constant volume, w = 0 (no PΔV work). ΔU = q + w = q + 0 = q. The system releases heat, so q = −400 J. Therefore ΔU = −400 J (NCERT Class 11 Chemistry Chapter 5, page 139).

Why B is wrong: B (+400 J) gets the sign of q wrong. The system releases heat, meaning q is negative (−400 J), not positive.

Why C is wrong: C (0 J) confuses constant volume with a cyclic process. At constant volume, ΔU = q_v, which is non-zero when heat is exchanged.

Why D is wrong: D is incorrect because at constant volume we don't need pressure. w = −P_ext ΔV = 0 since ΔV = 0, regardless of pressure. ΔU = q is fully determined.

Answer: C. Heat absorbed: q = +1000 J. Work done by system on surroundings = 400 J, so w = −400 J (chemistry convention: w = work on system). ΔU = q + w = 1000 + (−400) = 600 J (NCERT Class 11 Chemistry Chapter 5, page 139).

Why A is wrong: A (1400 J) adds q and |w| as if both are positive: 1000 + 400. This treats expansion work as work done on the system — the sign-convention trap. Expansion work by the system means w = −400 J.

Why B is wrong: B (1000 J) ignores the work term entirely (sets w = 0). Work was done during expansion, so w ≠ 0.

Why D is wrong: D (400 J) confuses ΔU with the work term alone, ignoring the heat absorbed.

Answer: B. In free expansion, P_ext = 0, so w = −P_ext ΔV = 0. For an ideal gas, internal energy depends only on temperature; since no work is done and the process is adiabatic (sudden expansion, no heat exchange), q = 0 and ΔU = 0. Temperature remains unchanged (NCERT Class 11 Chemistry Chapter 5, page 139).

Why A is wrong: A assumes the gas gains energy during expansion. With P_ext = 0, no work is done on or by the system, and free expansion into vacuum exchanges no heat.

Why C is wrong: C assumes w < 0 (expansion work). But expansion against zero external pressure means w = −P_ext ΔV = −(0)(ΔV) = 0, not negative.

Why D is wrong: D assumes heat is absorbed. Free expansion into vacuum is a rapid irreversible process with no thermal reservoir interaction; q = 0.