Physics · Thermodynamics · NEET
For NEET, use Q = dU + W. Here W is the work done BY the gas. If you rearrange it, you get dU = Q - W, which is the same equation written differently. Both are correct. The confusion comes from chemistry, where the formula is dU = Q - W but W there means work done ON the gas. In physics (NCERT), W is work done BY the gas, so Q = dU + W. Pick the physics form and stick to it.
W is the work done BY the gas on its surroundings. So when the gas expands, W is positive (it pushes the piston out). When the gas is compressed, W is negative (the surroundings push the piston in, so the gas does negative work). Always check: gas expands means +W, gas compressed means -W.
Heat is a signed quantity. If heat is given TO the gas (absorbed), Q is positive. If heat is released or rejected BY the gas, Q is negative. So a gas that rejects 200 J of heat has Q = -200 J. Put this signed value directly into Q = dU + W.
Rearrange the first law: dU = Q - W. Put in Q with its sign (positive if absorbed) and W with its sign (positive if the gas expands). Example: gas absorbs 500 J and does 200 J of work, so dU = 500 - 200 = 300 J. The internal energy increases by 300 J.
Only the sign of W changes. When the gas is compressed, work is done ON the gas, so the work done BY the gas is negative (W < 0). Then dU = Q - W becomes dU = Q - (negative) = Q + |W|, so the internal energy can rise even without heat. Everything else in the method stays the same.
At a certain temperature, during a process, 500 J is absorbed by the system and 200 J of work is done by the system. The change in internal energy of the system is
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, the rate at which internal energy increases is
0.1 g of water at 100 C requires 54 cal to convert into steam at 100 C at normal pressure (1.013 x 10^5 N/m^2). If the volume of steam produced is 167.1 cc, the change in internal energy is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Q = dU + W, where Q is the heat given to the gas, dU is the change in internal energy, and W is the work done by the gas. Rearranged, dU = Q - W. This is just the law of conservation of energy for a gas.
Use joules for the final answer. If heat is given in calories, convert using 1 cal = 4.18 J (some questions use 4.2 J). Work from P dV comes out in joules if pressure is in N/m^2 and volume in m^3, so keep all units in SI.
Q is positive when heat is absorbed by the gas and negative when heat is released. W is positive when the gas expands (does work on surroundings) and negative when the gas is compressed. Decide these two signs first, then just plug numbers in.
Q = dU + W always holds. What changes from process to process is how much of Q goes into dU versus W. For example, in an isochoric (constant volume) process W = 0 so Q = dU, and in an isothermal process dU = 0 so Q = W. The equation itself never changes.
Yes. If a gas is compressed with Q = 0 (adiabatic), then dU = -W. Since compression makes W negative, dU becomes positive, so the internal energy and temperature rise. This is why a bicycle pump gets hot when you press it fast.
