Question

Газ занимал V=1,5 м.3 при Т=24°С, Р=5кПа, уменьшили объем до 1м.куб, и Т=21°С, какое стало давление?

Answer 1

Ответ:

7,42 кПа

Объяснение:
24°=297 К
21°=294 К
5 кПа=5000 Па

p1V1/T1 = p2V2/T2
p1V1T2 = p2V2T1
p2=p1V1T2 / V2T1
p2=5000•1,5•294/1•297=2205000/297=7424 Па≈7,42 кПа

Answer 2

Calculation of the New Pressure

To calculate the new pressure after the volume and temperature change, we can use the ideal gas law equation:

P1V1/T1 = P2V2/T2

Where: - P1 is the initial pressure (5 kPa) - V1 is the initial volume (1.5 m^3) - T1 is the initial temperature (24°C) - P2 is the final pressure (unknown) - V2 is the final volume (1 m^3) - T2 is the final temperature (21°C)

Let's substitute the given values into the equation and solve for P2:

P1V1/T1 = P2V2/T2

5 kPa * 1.5 m^3 / (24°C + 273.15) = P2 * 1 m^3 / (21°C + 273.15)

Simplifying the equation:

7.5 kPa m^3 / 297.15 K = P2 * 1 m^3 / 294.15 K

Cross-multiplying:

7.5 kPa m^3 * 294.15 K = P2 * 1 m^3 * 297.15 K

Dividing both sides by 1 m^3 * 294.15 K:

P2 = (7.5 kPa m^3 * 294.15 K) / (1 m^3 * 297.15 K)

Calculating the value:

P2 ≈ 7.47 kPa

Therefore, the new pressure after reducing the volume to 1 m^3 and the temperature to 21°C is approximately 7.47 kPa.

Please note that the provided search results did not contain relevant information for this specific calculation. The calculation was performed using the ideal gas law equation.