A given quantity of gas is taken from A to C in two ways; a) directly from A → C along a straight line and b) in two steps, from A → B and then from B → C. Work done and heat absorbed along the direct path A→C is 200 J and 280 J respectively. If the work done along A→B→C is 80 J, then heat absorbed along this path is,
160 J
Internal energy is a state function, so its change between two states is the same for every path. The first law of thermodynamics links it to heat absorbed and work done by the gas: \(\Delta U = Q - W\).
Step 1 — Direct path A→C: \(\Delta U = 280 - 200 = 80\ \text{J}\).
Step 2 — Path A→B→C has the same end states, so \(\Delta U = 80\ \text{J}\) again, with \(W = 80\ \text{J}\).
Step 3 — Heat absorbed: \(Q = \Delta U + W = 80 + 80 = 160\ \text{J}\).
A value of 80 J equals only the change in internal energy (or only the work), leaving out the other term. Zero heat would require the internal energy to fall by 80 J, and 120 J does not satisfy the first law with the given work.
160 J option is correct.
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