Question

Определите температуру газа, если средняя кинетическая энер­гия хаотического движения его молекул равна 5,6 ×10 - 21 Дж. При какой температуре будет находиться газ, если средняя кинетическая энергия молекул уменьшится на 20%?

Answer 1

E=3/2kT отсюда Т= 2E/3k, где к=1,38*10^-38
E2=E1-0.2E1=0.8E1= 4.48*10^-21
аналогично находится Т2=2Е(2)/3к

Answer 2

Calculation of Gas Temperature

To determine the temperature of the gas, we can use the equation that relates the average kinetic energy of the gas molecules to the temperature. The equation is given by:

E_avg = (3/2) * k * T

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

We are given that the average kinetic energy of the gas molecules is 5.6 × 10^-21 J. We need to find the temperature of the gas when the average kinetic energy decreases by 20%.

Let's solve for the temperature using the given information.

1. Calculate the initial temperature (T1) using the given average kinetic energy (E_avg1): - E_avg1 = 5.6 × 10^-21 J - k = 1.38 × 10^-23 J/K

Using the equation E_avg1 = (3/2) * k * T1, we can solve for T1:

T1 = E_avg1 / [(3/2) * k]

2. Calculate the final temperature (T2) when the average kinetic energy decreases by 20%: - E_avg2 = 0.8 * E_avg1 (20% decrease)

Using the equation E_avg2 = (3/2) * k * T2, we can solve for T2:

T2 = E_avg2 / [(3/2) * k]

Let's calculate the values of T1 and T2:

1. Calculate T1: - E_avg1 = 5.6 × 10^-21 J - k = 1.38 × 10^-23 J/K

T1 = (5.6 × 10^-21 J) / [(3/2) * (1.38 × 10^-23 J/K)]

2. Calculate T2: - E_avg2 = 0.8 * E_avg1 - k = 1.38 × 10^-23 J/K

T2 = (0.8 * 5.6 × 10^-21 J) / [(3/2) * (1.38 × 10^-23 J/K)]

Now, let's calculate the values of T1 and T2 using the above equations.

Calculation:

1. Calculate T1: - E_avg1 = 5.6 × 10^-21 J - k = 1.38 × 10^-23 J/K

T1 = (5.6 × 10^-21 J) / [(3/2) * (1.38 × 10^-23 J/K)]

Using a calculator, we find that T1 ≈ 2.03 × 10^2 K.

2. Calculate T2: - E_avg2 = 0.8 * E_avg1 - k = 1.38 × 10^-23 J/K

T2 = (0.8 * 5.6 × 10^-21 J) / [(3/2) * (1.38 × 10^-23 J/K)]

Using a calculator, we find that T2 ≈ 1.63 × 10^2 K.

Therefore, the initial temperature of the gas (T1) is approximately 2.03 × 10^2 K, and the final temperature of the gas (T2) when the average kinetic energy decreases by 20% is approximately 1.63 × 10^2 K.

Please note that the above calculations are based on the given information and the assumption that the gas follows the ideal gas law and obeys the laws of thermodynamics.