Question

При понижении температуры газа до 200 К средняя кинетическая энергия хаотического движения молекул уменьшилась на 2*10^-21 Дж. Какой была энергия при первоначальной температуре? Определить эту первоначальную температуру.

Answer 1

Кинетическая энергия движения молекул равна Ek=3/2*kT если она изменилась, то   Ek-Eko=3/2*k(200-To)   To=200-2(Ek-Eko)/(3k)=200-(2*2*10^-21)/(3*1.38*10^-23)=66.6 K

Answer 2

Calculation of Initial Energy

To determine the initial energy at the original temperature, we can use the equation for the average kinetic energy of gas molecules:

E = (3/2) * k * T

Where: - E is the average kinetic energy of the gas molecules - k is the Boltzmann constant (1.38 * 10^-23 J/K) - T is the temperature in Kelvin

Given that the average kinetic energy decreased by 2 * 10^-21 J when the temperature decreased to 200 K, we can set up the following equation:

(3/2) * k * T_initial - (3/2) * k * 200 K = 2 * 10^-21 J

Now we can solve for T_initial, the initial temperature.

Let's calculate it step by step:

1. Multiply both sides of the equation by 2/3 to isolate T_initial:

k * T_initial - k * 200 K = (2/3) * 2 * 10^-21 J

2. Simplify the equation:

k * T_initial - k * 200 K = (4/3) * 10^-21 J

3. Add k * 200 K to both sides of the equation:

k * T_initial = k * 200 K + (4/3) * 10^-21 J

4. Divide both sides of the equation by k:

T_initial = (k * 200 K + (4/3) * 10^-21 J) / k

Now we can substitute the value of the Boltzmann constant (k = 1.38 * 10^-23 J/K) and calculate the initial temperature.

Let's calculate it:

T_initial = (1.38 * 10^-23 J/K * 200 K + (4/3) * 10^-21 J) / (1.38 * 10^-23 J/K)

T_initial ≈ 201.45 K

Therefore, the initial temperature was approximately 201.45 Kelvin.