Answer:
the pressure at the end of the combustion is 2.68 MPa
Solution:
As per the question:
Initial Pressure, P = 0.95\ MPa
Temperature before combustion, [tex]T = 425^{\circ}C[/tex] = 273 + 425 = 698 K
Temperature after combustion, [tex]T' = 1700^{\circ}C[/tex] = 1973 K
Now,
To calculate the pressure at the end of combustion, P':
By using the Pressure-Temperature relation from Gay- Lussac's law:
[tex]\frac{P}{T} = \frac{P'}{T'}[/tex]
[tex]P' = \frac{P}{T}\times T'[/tex]
[tex]P' = \frac{0.95\times 10^{6}}{698}\times 1973 = 2.68\times 10^{6}\ Pa = 2.68\ MPa[/tex]