An ideal gas undergoes an expansion from the initial state described by Pi, Vi, T to a final state described by Pf, Vf, T in (a) a process at the constant external pressure Pf, and (b) in a reversible process. Derive expressions for the largest mass that can be lifted through a height h in the surroundings in these processes.

Respuesta :

Answer:

A) m = [tex]\frac{p_f(V_f-V_i)}{gh}[/tex]

B) m =  [tex]- n R T (ln \frac{V_f }{V_i})\frac{1}{gh}[/tex]

Explanation:

A) a process at the constant external pressure Pf,

We Know that

W = mgh-------------------------(1)

W=[tex]p \Delta V[/tex]---------------------------(2)

Equating (1) and(2)

mgh =[tex]p \Delta V[/tex]

=[tex]p_f (V_f-V_i)[/tex]

Therefore m = [tex]\frac{p_f(V_f-V_i)}{gh}[/tex]

B)In a irreversible process

W = mgh----------------------(3)

[tex]\mathrm{W}=-\int_{V_{i}}^{V_{f}} \mathrm{P} d V[/tex]---------------------(4)

PV =nRT

P =[tex]\frac{nRT}{V}[/tex]

[tex]\mathrm{W}=-\int_{V_{i}}^{V_{f}} \frac{\mathrm{nRT}}{V} d V[/tex]

W =[tex]-n R T \int_{V_{i}}^{V_{f}} \frac{d V}{V}[/tex]

W= [tex]- n R T (ln V_i- ln V_f)[/tex]  

W=[tex]- n R T (ln V_f- ln V_i)[/tex]

W=[tex]- n R T( ln \frac{V_f }{ V_i})[/tex]

From EQ(3)

mgh= [tex]- n R T( ln \frac{V_f }{ V_i})[/tex]

m =  [tex]- n R T (ln \frac{V_f }{V_i})\frac{1}{gh}[/tex]

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