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]