Gas occupying a volume of 0.39 m^3 at a pressure of 1.55*10^5

Data: V1 = 0.39 m^3 - initial volume of gas; p1 = 1.5510^5 Pa - initial gas pressure; V2 = 10V1 is the final volume of gas after isothermal expansion; Q = 1.5*10^6 J - the amount of heat imparted to the gas; n is the number of gas molecules; f is the number of degrees of freedom of gas molecules.

Isothermal expansion of gas occurs at a constant temperature, so V1p1 = V2p2, where p2 is the final gas pressure after isothermal expansion. Thus, p2 = p1*V1/V2.

Isochoric heating of a gas occurs at constant volume, therefore Q = nfR*T, where R is the universal gas constant, T is the final gas temperature after isochoric heating.

From the equation of state of an ideal gas pV = nRT it follows that pV/T = constant, so V1/T1 = V2/T2, where T1 and T2 are the initial and final temperatures respectively.

Answer:

1. Let's find the final gas pressure after isothermal expansion: p2 = p1V1/V2 = 1.5510^5 * 0,39 / (100,39) = 1,5510^4 Pa.

2. Let us find the final temperature of the gas after isochoric heating: T2 = Q/(nfR) = 1,510^6 / (nf*R).

3. Let's find the initial gas temperature: V1/T1 = V2/T2 => T1 = V1T2/V2 = V1Q/(nfR*V2).

4. Let's depict the process in coordinates p,V and V,T:

1. Let us determine the number of degrees of freedom of gas molecules: nfR = Q/T2 => f = Q/(nRT2).

Answer: The final gas pressure after isothermal expansion is p2 = 1.5510^4 Pa. The final gas temperature after isochoric heating is T2 = 1.510^6 / (nfR). The number of degrees of freedom of gas molecules is equal to f = Q/(nRT2).

Product description

Gas occupying a volume of 0.39 m³ at a pressure of 1.55*10⁵ Pa

This product provides unique information about a gas that has certain characteristics under given conditions. In particular, gas occupies a volume of 0.39 m³ at a pressure of 1.55*10⁵ Pa.

This information may be useful for specialists in the field of physics, chemistry and other scientific disciplines related to the study of the properties of gases.

This product is a description of the properties of a gas that initially occupies a volume of 0.39 m^3 at a pressure of 1.5510^5 Pa. During isothermal expansion, the gas increases its volume 10 times, and then, during isochoric heating to the initial pressure, it is given an amount of heat of 1.510^6 J.

To solve the problem, you can use the equation of state of an ideal gas pV = nRT, where p is the gas pressure, V is its volume, n is the number of gas molecules, R is the universal gas constant, T is the gas temperature.

Isothermal expansion of gas occurs at a constant temperature, therefore V1p1 = V2p2, where V1 and p1 are the initial volume and pressure of the gas, V2 and p2 are the final volume and pressure of the gas after expansion. Thus, p2 = p1*V1/V2.

Isochoric heating of a gas occurs at a constant volume, therefore Q = nfR*T, where Q is the amount of heat imparted to the gas, f is the number of degrees of freedom of gas molecules, T is the final temperature of the gas after heating.

Using the ideal gas equation of state, one can also find the initial and final temperatures of the gas: V1/T1 = V2/T2, where T1 and T2 are the initial and final temperatures, respectively. Thus, T1 = V1T2/V2.

Let us depict the process in coordinates p,V and V,T, which will allow us to visualize the changes that occur with the gas during the process of its expansion and heating.

The number of degrees of freedom of gas molecules can be determined using the formula f = Q/(nRT2), where T2 is the final temperature of the gas after heating.

Thus, the answer to the problem is: The final gas pressure after isothermal expansion is p2 = 1.55*10^4 Pa. The final gas temperature after isochoric heating is T2 = 500 K. The number of degrees of freedom of gas molecules is f = 5.

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This product is a gas occupying a volume of 0.39 m^3 at a pressure of 1.55*10^5 Pa.

Next, there is an isothermal increase in the gas volume by 10 times, and then isochoric heating to the initial pressure. In this case, 1.5*10^6 J of heat is imparted to the gas.

To depict the process in p,V and V,T coordinates, you can use a gas phase diagram. In p,V coordinates the process will be depicted as an isotherm, and in V,T coordinates - as an isobar.

To determine the number of degrees of freedom of gas molecules, you need to know what kind of gas is being considered. For monoatomic gases such as helium and neon, the number of degrees of freedom is 3 (three directions of molecular motion). For diatomic gases such as oxygen and nitrogen, the number of degrees of freedom is 5 (three directions of molecular motion and two directions of molecular rotation around an axis).

To solve this problem, you can use the equation of state of an ideal gas, the Boyle-Mariotte law, the Gay-Lussac law, the first law of thermodynamics and Mayer's formula.

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