Question

An ideal gas at 20 ∘C consists of 2.2×1022 atoms. 8.0 J of thermal energy are removed from the gas. What is the new temperature in ∘C?

Answer 1

Write the formulka for thermal energy. \(\Delta E_{t h}=\frac{3}{2} N k_{B} \Delta T\)

(1)

From the equaiton (1), \(\Delta T=\frac{2}{3} \frac{\Delta E_{t h}}{N k_{B}}\)

\(\Delta T=\frac{2}{3}\left(\frac{-8}{\left(2.2 \times 10^{22}\right)\left(1.38 \times 10^{-23}\right)}\right)\)

\(\Delta T=-17.57^{\circ} \mathrm{C}\)

Calculate the new temparature. \(T_{f}=T_{i}+\Delta T\)

\(T_{f}=20^{\circ} \mathrm{C}-17.57^{\circ} \mathrm{C}\)

\(T_{f}=2.43^{\circ} \mathrm{C}\)