Is n=1 (degrees of freedom) for monatomic?

Dayanara Terry
2022-07-15
Answered

Is n=1 (degrees of freedom) for monatomic?

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Isla Klein

Answered 2022-07-16
Author has **12** answers

In the ideal gas law

$PV=nRT$

n is not the degrees of freedom. Instead n is the number of molecules (measured in mol). You can see this also from the unit of the gas constant

$R=8.314{\text{m}}^{3}\text{Pa}{\text{K}}^{-1}{\text{mol}}^{-1}$

$PV=nRT$

n is not the degrees of freedom. Instead n is the number of molecules (measured in mol). You can see this also from the unit of the gas constant

$R=8.314{\text{m}}^{3}\text{Pa}{\text{K}}^{-1}{\text{mol}}^{-1}$

asked 2022-08-13

When we are dealing with a gaseous thermodynamic system, in books it's written that state of the system can be described by only two independent variables from the three (p,V,T) . But it's not written why or how? Why we have to choose only two independent variables, why not more or less? I've gone through some answers but still unable to understand.

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In thermodynamics, the enthalpy of a system is defined as the sum of the internal energy of the system and the product of its pressure and volume. Since it is just a combination of other state properties of the system, why need we define it at all?

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Scientists say that it is impossible to reach a temperature of zero kelvin, because the atoms will stop moving, and the volume of the substance will become zero, But we have reached the pico kelvin temperature, which is almost zero, and there is no difference between it and zero, so why do they say it is impossible? and The laws of physics still work at picokelvin.

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How to justify Mechanical Works doesn't involve heat variations?

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asked 2022-10-16

Could anyone give some definite proof for this?

$\int \frac{{Q}_{rev}}{T}=\mathrm{\Delta}({k}_{B}\mathrm{ln}\mathrm{\Omega})=\mathrm{\Delta}S$

$\int \frac{{Q}_{rev}}{T}=\mathrm{\Delta}({k}_{B}\mathrm{ln}\mathrm{\Omega})=\mathrm{\Delta}S$

asked 2022-11-07

I have to prove, considering the limits of high and low temperatures, duality equation:

$Z=q{e}^{2NK}F\left({e}^{-K}\right)={q}^{-N}{({e}^{K}+q-1)}^{2N}F\left(\frac{{e}^{K}-1}{{e}^{K}+q-1}\right)$

for square lattice with

$Z=\sum _{{\sigma}_{1},{\sigma}_{N}}\prod _{<ij>}{e}^{K{\delta}_{{\sigma}_{i}{\sigma}_{j}}}$

$Z=q{e}^{2NK}F\left({e}^{-K}\right)={q}^{-N}{({e}^{K}+q-1)}^{2N}F\left(\frac{{e}^{K}-1}{{e}^{K}+q-1}\right)$

for square lattice with

$Z=\sum _{{\sigma}_{1},{\sigma}_{N}}\prod _{<ij>}{e}^{K{\delta}_{{\sigma}_{i}{\sigma}_{j}}}$