Heat at constant volume (monatomic gas)

Problem

$2\ \text{mol}$ of monatomic gas ($f = 3$) are heated from $T_A = 300\ \text{K}$ to $T_B = 350\ \text{K}$ in a rigid container. How much heat is needed?

Solution

Constant volume. The container is rigid, so $L = 0$ and all the heat goes into internal energy: $Q = \Delta E_\text{int} = \frac{f}{2}\, n R\, \Delta T = \frac{3}{2}\, n R\, \Delta T$

Data. $f = 3$; $n = 2\ \text{mol}$; $\Delta T = 50\ \text{K}$.

Numerical substitution. $Q = \frac{3}{2}\cdot 2\cdot 8{,}31\cdot 50 = 1246{,}5\ \text{J}$

$\ev{Q \approx 1{,}25\cdot 10^3\ \text{J}}$

Links

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