Question

Идеальный одноатомный газ нагревается от температуры Т1 = 300К до температуры Т2 = 400К таким образом , что его температура меняется линейно с изменением давления от Р1 = 1атм до Р2 = 2атм . Найти отношение конечной и начальной теплоёмкостей газа в этом процессе

Answer 1

Решение во вложении
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Answer 2

Calculation of the Ratio of Specific Heats for an Ideal Monoatomic Gas

To find the ratio of the final and initial specific heats of an ideal monoatomic gas, we can use the relationship between pressure, volume, and temperature for an adiabatic process. In this case, the temperature of the gas changes linearly with the change in pressure.

Let's denote the initial temperature as T1 = 300 K, the final temperature as T2 = 400 K, the initial pressure as P1 = 1 atm, and the final pressure as P2 = 2 atm.

The relationship between pressure, volume, and temperature for an adiabatic process is given by the equation:

P1 * V1^γ = P2 * V2^γ,

where γ is the ratio of the specific heats of the gas.

To find the ratio of specific heats, we need to express the volume in terms of pressure and temperature. For an ideal gas, the equation of state is given by:

P * V = n * R * T,

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

Since we are considering an ideal monoatomic gas, the number of moles of gas can be expressed in terms of the mass of the gas and the molar mass:

n = m / M,

where m is the mass of the gas and M is the molar mass.

Combining these equations, we can express the volume in terms of pressure and temperature:

V = (n * R * T) / P.

Substituting this expression for V into the equation for the adiabatic process, we get:

(P1 * [(n * R * T1) / P1])^γ = (P2 * [(n * R * T2) / P2])^γ.

Simplifying this equation, we find:

T1^γ = T2^γ.

Taking the logarithm of both sides of the equation, we get:

γ * ln(T1) = γ * ln(T2).

Dividing both sides of the equation by γ, we find:

ln(T1) = ln(T2).

Since the natural logarithm of a number is equal to the logarithm of that number to any base, we can choose any base for the logarithm. Let's choose the base 10.

Taking the base 10 logarithm of both sides of the equation, we get:

log10(T1) = log10(T2).

Finally, dividing both sides of the equation by log10(T1), we find:

log10(T2) / log10(T1) = 1.

Therefore, the ratio of the final and initial specific heats of the gas in this process is 1.

Please note that the ratio of specific heats for an ideal monoatomic gas is always equal to 1, regardless of the process.