Answer:
The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have
The temperature T = 602.64K and the final pressure P = 110.24MPa
Explanation:
Numbeer of moles of gas = 3.00 mol
initial volume = 230.8×10−6 m3
final volume = 133.4×10−6 m3 .
released energy = 8240 J
Temperature = Constant = T
Pressure = [tex]p_{f}[/tex] =unknown
From the relation the combined ideal gas law, PV = nRT
Where R = 8.314 4621.JK−1mol−1
we have The release energy from compression P1V1 -P2V2
-qrev = -nRTln([tex]\frac{V_{2} }{V_{1} }[/tex]) = 8240J
n = 3
Hence -nRTln([tex]\frac{V_{2} }{V_{1} }[/tex]) = 3×8.314 462×ln[tex](\frac{133.4}{230.8})[/tex] × T= -8240 J
or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K
The Final pressure is given by
PV = n×R×T from where we have V = final volume thus
P = (n×R×T)/V = (3×8.134×602.64)÷(133.4×[tex]10^{-6}[/tex]) = 110237041.1 N/[tex]m^{2}[/tex] = 110.237MPa
