Thermodynamics: Speed-Solving Techniques (2)

Question

An ideal monatomic gas undergoes an isothermal expansion at $T = 300 \, \text{K}$ from $V_1 = 2 \, \text{L}$ to $V_2 = 6 \, \text{L}$. Find the heat absorbed for $n = 0.5$ mol — in 60 seconds.

Solution — Step by Step

Why This Works

The shortcut $\Delta U = 0$ for isothermal works for any ideal gas — monatomic, diatomic, polyatomic. We do not need the heat capacity at all.

That makes isothermal questions among the fastest in thermodynamics: one formula, one log, done.

Alternative Method — Direct Integration

We could integrate $W = \int P \, dV$ with $P = nRT/V$ from $V_1$ to $V_2$. That gives the same formula. But for a timed paper, memorising the result saves 30+ seconds.

Common Mistake

Students often write $Q = nC_V \Delta T$ for isothermal — but $\Delta T = 0$, giving $Q = 0$, which is wrong. The error: applying $Q = nC_V \Delta T$ is only valid for constant-volume processes.

For NEET, the heat-and-work split for each thermodynamic process (isothermal, isobaric, isochoric, adiabatic) is a 4-mark scoring topic almost every year.