An ideal gas at 0∘C consists of 1.0*1023 atoms. 20 J of thermal energy are added to the gas. Part A What is the new temperature in ∘C?

Answer: New temperature T2 = 9.66°C

Explanation:

Given;

Energy added ∆E = 20J

Initial temperature T1 = 0°C

Number of atom = 1.0*10^23

Using the energy equation.

∆E = (3/2) nR∆T

∆T = (2/3) ∆E/nR

Where;

R = universal gas constant = 8.314J/mol-K

n = number of moles = N/Nt

Nt = number of atoms in one mole of gas = 6.022*10^23

n = 1.0 * 10^23/6.022*10^23 = 0.166

∆T = (2/3) * 20 / (0.166*8.314)

∆T = 9.66

T2 = T1+∆T = 0 + 9.66

T2 = 9.66°C