Question

кислород при температуре 27 градусов и давление 5*10^5 Па изохорно охладили после чего давление стало равно 4*10^5 Па. До какой температуры охладили кислород

Answer 1

Дано:

T1=27 град. Цельсия=27+273=300 К.

P1=5*10^5 Па.

P2=4*10^5 Па.

T2=?
________

Решение:
Процесс происходящий при неизменном объеме называется изохорным. В нашей задаче как раз такой и описывается.

Записываем закон Менделеева - Клапейрона для двух случаев:
$></p><p>Считаем: T2=(300*4*10^5)/(5*10^5)=240 К=-33 град. Цельсия.</p><p>Ответ: T2=240 К.</p></div> <div class=$ 0 0

Answer 2

Calculation of the Final Temperature of Oxygen

To determine the final temperature to which oxygen was cooled, we can use the ideal gas law, which states that the pressure, volume, and temperature of an ideal gas are related by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, the process is isochoric, meaning the volume remains constant. Therefore, we can simplify the ideal gas law equation to P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

Given: - Initial temperature (T1) = 27 degrees Celsius = 27 + 273.15 Kelvin - Initial pressure (P1) = 5 * 10^5 Pa - Final pressure (P2) = 4 * 10^5 Pa

We can rearrange the equation to solve for the final temperature (T2):

T2 = (P2 * T1) / P1

Substituting the given values:

T2 = (4 * 10^5 * (27 + 273.15)) / (5 * 10^5)

Calculating the value:

T2 = (4 * 10^5 * 300.15) / (5 * 10^5)

T2 = 240.12 Kelvin

Therefore, the oxygen was cooled to a final temperature of approximately 240.12 Kelvin.