Respuesta :
Answer:
The final pressure is 148 kPa
Solution:
As per the question:
Initial Pressure of the gas, P = 111 kPa
Initial Volume, V = 5.00 l
Now,
In order to calculate the final pressure of the system:
At constant pressure, the work done is 'W'
If the absolute initial temperature be T
When the absolute temperature is doubled, T' =2T
Total work done in the processes, W = [tex]\frac{W}{2}[/tex]
Now,
From the ideal gas eqn:
PV = nRT
PV ∝ T (1)
In constant pressure process, work done is given by:
W = [tex]P\Delta V[/tex] (2)
Thus from eqn (1) and (2):
[tex]\frac{V}{V'}= \frac{T}{T'}[/tex]
In the first process, temperature is doubled:
[tex]\frac{V}{V'}= \frac{T}{2T}[/tex]
V = 2V'
Thus
[tex]V' = \frac{V}{2}[/tex]
In the process, the change in volume is [tex]\frac{V}{2}[/tex] given by:
[tex]V_{t} = V + \frac{V}{2} = \frac{3V}{2}[/tex]
Now,
[tex]\frac{PV}{P'V'} = \frac{T}{T'}[/tex]
[tex]\frac{PV}{P'\frac{3V}{2}} = \frac{T}{2T}[/tex]
[tex]\frac{P}{P'\frac{3}{2}} = \frac{1}{2}[/tex]
[tex]P' = \frac{4}{3}P[/tex]
[tex]P' = \frac{4}{3}\times 111 = 148\ kPa[/tex]
