Question

An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 230.8 × 10 − 6 m 3 230.8×10−6 m3 to a final volume of 133.4 × 10 − 6 m 3 133.4×10−6 m3 . If 8240 J 8240 J is released by the gas during this process, what are the temperature T T and the final pressure p f pf of the gas?

Answer 1

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 =  $p_{f}$ =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($\frac{V_{2} }{V_{1} }$) = 8240J

n = 3

Hence -nRTln($\frac{V_{2} }{V_{1} }$) =  3×8.314 462×ln$(\frac{133.4}{230.8})$ × 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×$10^{-6}$) = 110237041.1 N/$m^{2}$ = 110.237MPa