Question

Термодинамика, 10 класс Дано:Идеальный газ (аргон) массой 0,2кг при температуре Т1 и давлении р (кПа) занимает объем 0,073 м^3. Внутренняя энергия газа 75.4 кДж. В изобарном процессе газу сообщили ( отвели ) количество теплоты Q . При этом газом совершена работа А, приращение внутренней энергии ΔU, конечная температура 350К.Найти:p, T1, A, ΔU, Q - ?

Answer 1

Ответ:

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Answer 2

Given Information:

- Mass of the ideal gas (argon): 0.2 kg - Initial temperature: T1 - Initial pressure: p (in kPa) - Initial volume: 0.073 m^3 - Internal energy of the gas: 75.4 kJ - Final temperature: 350 K

Finding the Initial Pressure (p):

To find the initial pressure of the gas, we can use the ideal gas law equation:

pV = nRT

Where: - p is the pressure - V is the volume - n is the number of moles of gas - R is the ideal gas constant - T is the temperature

Since we are given the mass of the gas (0.2 kg) and the molar mass of argon is approximately 39.95 g/mol, we can calculate the number of moles (n) using the formula:

n = mass / molar mass

Let's calculate the number of moles (n) first:

n = 0.2 kg / (39.95 g/mol) = 0.005006 mol

Now, we can rearrange the ideal gas law equation to solve for the initial pressure (p):

p = (nRT) / V

Substituting the known values:

p = (0.005006 mol * R * T1) / 0.073 m^3

Since we don't have the value of T1, we cannot calculate the exact value of p without additional information.

Finding the Work Done (A):

In an isobaric process, the pressure remains constant. The work done by the gas can be calculated using the formula:

A = p * ΔV

Where: - A is the work done - p is the pressure - ΔV is the change in volume

Since the process is isobaric, the pressure remains constant, and the change in volume (ΔV) can be calculated as:

ΔV = V2 - V1

Substituting the known values:

ΔV = 0.073 m^3 - 0.073 m^3 = 0 m^3

Therefore, the work done (A) is zero, as there is no change in volume.

Finding the Change in Internal Energy (ΔU):

The change in internal energy (ΔU) can be calculated using the formula:

ΔU = Q - A

Where: - ΔU is the change in internal energy - Q is the heat added to the system - A is the work done by the system

Since we know that the work done (A) is zero, the change in internal energy (ΔU) is equal to the heat added to the system (Q).

Therefore, ΔU = Q = 75.4 kJ.

Finding the Heat Added (Q):

The heat added to the system (Q) can be calculated using the formula:

Q = ΔU + A

Since we know that the work done (A) is zero, the heat added (Q) is equal to the change in internal energy (ΔU).

Therefore, Q = ΔU = 75.4 kJ.

Summary of Results:

- Initial pressure (p): Cannot be determined without the value of T1. - Work done (A): 0 J (no change in volume) - Change in internal energy (ΔU): 75.4 kJ - Heat added (Q): 75.4 kJ

Please note that the value of the initial pressure (p) cannot be determined without the value of T1.