Isothermal and adiabatic processes

Isothermal — heat flows, temperature does not. A process is isothermal when the gas stays at one fixed temperature $T$ throughout, which requires it to be in close, slow (quasi-static) contact with a large heat reservoir. For a fixed amount of ideal gas this is just Boyle's law:

where: $p$ is pressure, $V$ is volume, and the product stays fixed only because $T$ does not change. Since $T$ is constant, the internal energy of an ideal gas is unchanged, $\Delta U = 0$, so the first law of thermodynamics ($\Delta U = Q - W$) collapses to $Q = W$: every joule of heat taken in during expansion leaves as work done by the gas, and every joule of work done on the gas during compression leaves as heat rejected.

Adiabatic — no heat flows, temperature moves. A process is adiabatic when $Q=0$ throughout — either the system is thermally insulated, or it happens too fast for heat to cross the boundary. For a reversible adiabatic change in an ideal gas,

where: $\gamma = C_p/C_v$ is the ratio of the molar heat capacity at constant pressure to that at constant volume, always greater than $1$ for a real gas. The syllabus does not require deriving this relation from scratch — it is stated and used directly. With $Q=0$, the first law reduces to $\Delta U = -W$: any work the gas does during adiabatic expansion is paid for entirely out of its own internal energy, so the gas cools; any work done on the gas during adiabatic compression raises its internal energy, so it warms.

A useful corollary. Combining $pV^{\gamma}=\text{constant}$ with the ideal-gas law $pV=nRT$ (eliminating $p$) gives a direct temperature–volume relation for an adiabatic process:

where: $T$ is absolute temperature and $V$ is volume at any instant of the same adiabatic process. This is not a new law, just the same relation written in different variables, and it is the fastest route to a final temperature once $p_1V_1^\gamma=p_2V_2^\gamma$ has already been used for pressure. On a $p$–$V$ diagram both curves fall as $V$ increases, but the adiabatic curve through any point is steeper than the isothermal curve through the same point, because an expanding gas with no incoming heat must give up its own internal energy — pressure and temperature both drop faster than heat-compensated Boyle's-law cooling would allow.

A gas with $\gamma = 1.4$ undergoes a reversible adiabatic compression from $V_1 = 8.0\ \text{L}$ at $p_1 = 1.0\times10^5\ \text{Pa}$ to $V_2 = 1.0\ \text{L}$. Suppose it started at room temperature, $T_1 = 300\ \text{K}$. Find the final pressure $p_2$ and the final temperature $T_2$.

Step 1. Apply the adiabatic pressure relation: $p_1V_1^\gamma = p_2V_2^\gamma$, so $p_2 = p_1\left(\dfrac{V_1}{V_2}\right)^\gamma = (1.0\times10^5)(8.0)^{1.4}$.

Step 2. Evaluate $(8.0)^{1.4}$. Since $8^{0.4} = e^{0.4\ln 8} = e^{0.832} \approx 2.30$, we get $(8.0)^{1.4} \approx 8\times2.30 = 18.4$, giving $p_2 \approx (1.0\times10^5)(18.4) = 1.84\times10^6\ \text{Pa}$.

Step 3. Apply the temperature–volume relation to the same compression: $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$, so $T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1} = T_1(8.0)^{0.4}$. The factor $(8.0)^{0.4}\approx2.30$ was already computed in Step 2 — no fresh exponentiation is needed.

Step 4. Substitute $T_1 = 300\ \text{K}$: $T_2 \approx 300 \times 2.30 = 690\ \text{K}$. The gas is compressed to $p_2\approx1.84\times10^6\ \text{Pa}$ and heats from $300\ \text{K}$ to about $690\ \text{K}$ — more than double, purely from the work done on it, since no heat could escape to soften the temperature rise.

• Applying Boyle's law ($pV=\text{const}$) to an adiabatic process, or the adiabatic relation ($pV^\gamma=\text{const}$) to an isothermal process — the two describe genuinely different physical conditions and are never interchangeable.
• Assuming "adiabatic" means "no temperature change" — it is heat $Q$ that is zero, not $\Delta T$; temperature typically changes a great deal in an adiabatic process.
• Forgetting that $\gamma>1$ always, so at any shared point the adiabatic curve is steeper than the isothermal curve, by exactly the factor $\gamma$ shown above.
• Trying to derive $pV^\gamma=\text{const}$ itself from scratch in an exam answer — the syllabus explicitly does not require this; state and use the relation directly.

Ask the tutor: want to see how isothermal and adiabatic legs combine in a full Carnot cycle, or try more $pV^\gamma$ and $TV^{\gamma-1}$ practice with different $\gamma$ values?