Question about the thermodynamic temperature scale

[MatinSAR]

Homework Statement

Prove the equality of ideal gas and thermodynamics temperature for a specific gas.

Relevant Equations

$\dfrac {Q_1}{Q_2}= \dfrac {T_1}{T_2}$

My first problem is to find the absored and rejected heat. Can I say that it is equal to the work done in an isothermal proccess ($dQ=Pdv$)?

My reasoning : We have $dQ=C_V d\theta + Pdv$. For constant temperature it becomes :$$dQ=Pdv$$

 

[connectednatural]

The gas equation of state is expressed as P(v-b) = Rθ, where P is pressure, v is specific volume, b is a constant, R is the gas constant, and θ is temperature in Kelvin. The heat capacity, CV, depends only on temperature θ. To demonstrate θ = T, we use the Carnot cycle. In this cycle, efficiency (η) is given by 1 - Tc/Th, where Tc is the cold reservoir temperature and Th is the hot reservoir temperature.

In a Carnot cycle, efficiency is also expressed as 1 - Qc/Qh. Utilizing the first law of thermodynamics, we substitute Qc = CV(θc - θh) and Qh = CV(θh - θc) into the efficiency equation, solving for θ to find θ = T.

 

[MatinSAR]

The gas equation of state is expressed as P(v-b) = Rθ, where P is pressure, v is specific volume, b is a constant, R is the gas constant, and θ is temperature in Kelvin. The heat capacity, CV, depends only on temperature θ. To demonstrate θ = T, we use the Carnot cycle. In this cycle, efficiency (η) is given by 1 - Tc/Th, where Tc is the cold reservoir temperature and Th is the hot reservoir temperature.

In a Carnot cycle, efficiency is also expressed as 1 - Qc/Qh. Utilizing the first law of thermodynamics, we substitute Qc = CV(θc - θh) and Qh = CV(θh - θc) into the efficiency equation, solving for θ to find θ = T.

This is another method to solve. Thanks for your reply ...
What's your idea about what I've said?

Can I say that the absored or rejected heat is equal to the work done in an isothermal proccess ($dQ=Pdv$)? Because we have $dQ=C_V d\theta + Pdv$. For constant temperature it becomes :$$dQ=Pdv$$