A gas is characterised by the enthalpy H(S, p) = Cp T, where Cp is a constant (called heat capacity and defined in § 5.2), and by pV = NR T, where p is its pressure, V its volume, T its temperature and N the number of moles of gas. An adiabatic reversible compression brings the pressure from p1 to

Question

A gas is characterised by the enthalpy [latex]H(S, p) = C_p T[/latex], where [latex]C_p[/latex] is a constant (called heat capacity and defined in § 5.2), and by pV = N R T, where p is its pressure, V its volume, T its temperature and N the number of moles of gas. An adiabatic reversible compression brings the pressure from [latex]p_1 to p_2[/latex] where [latex]p_2 > p_1[/latex]. The initial temperature is [latex]T_1[/latex]. Determine the temperature [latex]T_2[/latex] at the end of the compression.

Accepted Answer

For a reversible adiabatic process, we have dS = 0. Then, the enthalpy differential is given by,

[latex] dH = C_p dT = T dS + V dp = V dp [/latex].

Since p V = N R T, it can be recast as,

[latex] \frac{dT}{T} = \frac{NR}{C_p} \frac{dp}{p} [/latex].

The integration of this relation from the initial state [latex](T_1, p_1) [/latex] to the final state [latex](T_2, p_2) [/latex] yields,

[latex]\ln \Bigl(\frac{T_{2}}{T_{1}}\Bigr) =\frac{NR}{C_{p}} \ln \Bigl(\frac{p_{2}}{p_{1}}\Bigr)[/latex].

The exponentiation of this equation yields the temperature at the end of the compression,

[latex]T_2 =T_1 \Bigl(\frac{p_2}{p_1}\Bigr)^{\frac{NR}{C_p} } [/latex].

Question 4.1: A gas is characterised by the enthalpy H(S, p) = Cp T, where...

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