Adiabatic compression

Problem

Two moles of monatomic gas ($\gamma = 5/3$) are adiabatically compressed from $V_1 = 4\ \text{L}$ to $V_2 = 1\ \text{L}$, starting from $T_1 = 300\ \text{K}$. Calculate the final temperature $T_2$. This is the same principle behind Diesel engines, where rapid compression heats the air until ignition.

Solution

Adiabatic law. In a reversible adiabatic process

$T V^{\gamma-1} = \text{const}$

so $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.

Isolating $T_2$. With $\gamma - 1 = \tfrac{5}{3}-1 = \tfrac{2}{3}$ and $V_1/V_2 = 4$:

$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300\cdot 4^{2/3} = 300\cdot 2{,}520 \approx 756\ \text{K}$

Interpretation. By compressing without exchanging heat, the work done on the gas accumulates as internal energy and the temperature rises sharply (from $300$ to about $756\ \text{K}$): this is what triggers spontaneous ignition in the Diesel cycle.

$\ev{T_2\approx 756\ \text{K}}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes Objects: Ideal gas