Question

работа совершенная идеальным одноатомным газом и адиабатном процессе при уменьшении его внутренней энергии на 20кДж​

Answer 1

Ответ:

Работа газа равна 20 кДж.

Объяснение:

ΔU = - 20 кДж

Процесс: адиабатный

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А - ? - работа газа

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По Первому закону термодинамики

Q = ΔU + A

Адиабатный процесс — это процесс, при котором не происходит теплообмена с внешней средой, то есть

Q = 0

Тогда работа газа

А = - ΔU = 20 кДж.

Таким образом, при уменьшении внутренней энергии, газ совершает положительную работу.

Answer 2

Working with an Ideal Monoatomic Gas in an Adiabatic Process

When working with an ideal monoatomic gas in an adiabatic process, we are considering a system where no heat is exchanged with the surroundings. In this case, the change in the internal energy of the gas is solely due to work done on or by the gas.

To find the change in the internal energy of the gas, we need to know the initial and final states of the gas and the work done on or by the gas during the process. In this case, we are given that the internal energy of the gas decreases by 20 kJ.

The internal energy of an ideal monoatomic gas is given by the equation:

ΔU = (3/2) nR ΔT

Where: - ΔU is the change in internal energy - n is the number of moles of gas - R is the gas constant - ΔT is the change in temperature

Since we are not given the number of moles of gas or the change in temperature, we cannot directly calculate the work done on or by the gas. However, we can provide some general information about adiabatic processes.

In an adiabatic process, the relationship between pressure (P) and volume (V) is given by:

P1 * V1^γ = P2 * V2^γ

Where: - P1 and P2 are the initial and final pressures, respectively - V1 and V2 are the initial and final volumes, respectively - γ is the heat capacity ratio, which is equal to the ratio of specific heat capacities at constant pressure (Cp) and constant volume (Cv)

For an ideal monoatomic gas, γ is equal to 5/3.

Unfortunately, without more specific information about the process or the gas, we cannot provide a more detailed analysis or calculate the exact values of pressure, volume, or work done.

If you have any additional information or specific questions, please let me know and I'll be happy to assist you further.