Nitrogen heated in a rigid cylinder

Problem

$20\ \text{mol}$ of nitrogen (diatomic) in a rigid container of $1\ \text{m}^3$ are heated from $T_A = 300\ \text{K}$ to $T_B = 320\ \text{K}$. How much heat must be supplied?

Solution

Setup. The container is rigid: the volume does not change, so the gas performs no work and the atmosphere is not involved either: $L_\text{gas} = 0, \qquad L_\text{atm} = 0$

First law. With $L_\text{gas} = L_\text{atm} = 0$ all the heat goes into internal energy: $Q = \Delta E_\text{int} = \frac{f}{2}\, n R\, (T_B - T_A)$

Data. Diatomic nitrogen $\Rightarrow f = 5$; $n = 20\ \text{mol}$; $\Delta T = T_B - T_A = 20\ \text{K}$.

Numerical substitution. $Q = \frac{5}{2}\cdot 20\cdot 8{,}31\cdot 20 = 4155\ \text{J}$

$\ev{Q \approx 4155\ \text{J}}$

Note: $Q$ does not depend on the volume of the container; there is no work because the volume does not change.

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Topics: Thermodynamics Concepts: First law of thermodynamics · Internal energy Objects: Ideal gas