Question

Кислород при температуре 270 С и давлении 5·105 Па изохорно охладили, после чего давление стало равно 4·105 Па. До какой температуры охладили кислород

Answer 1

Переводим в СИ Т1=270+273=543К
Используя закон Шарля
$\frac{P1}{T1} = \frac{P2}{T2}$
$\frac{5*10 ^{5} }{543} = \frac{4*10 ^{5} }{T2} T2= \frac{4*543}{5} =434.4K$
 или 161,4С

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 product of pressure and volume is directly proportional to the product of the number of moles of gas and the temperature. Mathematically, this can be expressed as:

P1 * V1 / T1 = P2 * V2 / T2

Where: - P1 is the initial pressure of the oxygen gas (5 * 10^5 Pa) - V1 is the initial volume of the oxygen gas (since the process is isochoric, the volume remains constant) - T1 is the initial temperature of the oxygen gas (270 °C + 273.15 = 543.15 K) - P2 is the final pressure of the oxygen gas (4 * 10^5 Pa) - V2 is the final volume of the oxygen gas (since the process is isochoric, the volume remains constant) - T2 is the final temperature of the oxygen gas (the value we need to find)

By rearranging the equation, we can solve for T2:

T2 = (P2 * V2 * T1) / (P1 * V1)

Substituting the given values:

T2 = (4 * 10^5 Pa * V2 * 543.15 K) / (5 * 10^5 Pa * V1)

Since the volume remains constant, we can cancel out V2 and V1:

T2 = (4 * 10^5 Pa * 543.15 K) / (5 * 10^5 Pa)

Now we can calculate the final temperature of the oxygen gas.

Calculation:

T2 = (4 * 10^5 Pa * 543.15 K) / (5 * 10^5 Pa)

T2 ≈ 434.52 K

Therefore, the oxygen gas was cooled to approximately 434.52 K.