Question

Помогите решить задачу по физике!! Отмечаю правильный ответ лучшим и даю 35 баллов!! В сосуде объемом 1,2 находится газ под давлением 4* 10^5 Па. Какой объем будет занимать этот газ при давлении 2* 10^5 Па? Температура газа постоянна

Answer 1

Ответ:

Объяснение:

Дано:

V₁ = 1,2 л = 1,2·10⁻³ м³

p₁ = 4·10⁵ Па

p₂ = 2·10⁵ Па

T - const

__________

V₂ - ?

Поскольку процесс изотермический, то воспользуемся законом Бойля-Мариотта:

p₁·V₁ = p₂·V₂

V₂ = p₁·V₁ / p₂·

V₂ = 4·10⁵·1,2·10⁻³ / (2·10⁵) = 2,4·10⁻³ м³    или     2,4 литра

Answer 2

Problem Statement:

You have a vessel with a volume of 1.2 and it contains gas at a pressure of 4*10^5 Pa. You want to find the volume that the gas will occupy when the pressure is reduced to 2*10^5 Pa. The temperature of the gas remains constant.

Solution:

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when the temperature is constant. Mathematically, Boyle's Law can be expressed as:

P1 * V1 = P2 * V2

Where: P1 = initial pressure V1 = initial volume P2 = final pressure V2 = final volume

In this case, we know the initial pressure (P1 = 4*10^5 Pa), the initial volume (V1 = 1.2), and the final pressure (P2 = 2*10^5 Pa). We need to find the final volume (V2).

Let's substitute the known values into Boyle's Law equation and solve for V2:

(4*10^5 Pa) * (1.2) = (2*10^5 Pa) * V2

Simplifying the equation:

V2 = (4*10^5 Pa * 1.2) / (2*10^5 Pa)

V2 = 2.4

Therefore, the final volume that the gas will occupy when the pressure is reduced to 2*10^5 Pa is 2.4.

Answer:

The final volume that the gas will occupy when the pressure is reduced to 2*10^5 Pa is 2.4.