Answer:
The answer to this question can be given as:
The initial pressure P= 30 lbf/in2
The initial temperature T1=80°F.
The initial volume=final volume V= 1.6 ft3.
Final temperature T2=500°F
Heat transfer Q=-2.6 Btu.
pressure (P) (lbf/in2) change into lbf/ft2
P=30 lbf/in2=4320lbf/ft2.
Temperature (T1) °F change into °K
temperature T1=80°F=539.67 °K
Ideal gas formula:
M=PV/RT1. where T1=initial temperature.
M=4320*1.6/35.114*539.67.
M=0.364749 lb Mass at Carbon dioxide (CO2).
DQ=DU+W second law.
-2.6=MCV(T2-T1)+ W
-2.6-(0.364749*1.6(500-80))=W
W=-27.111 b+u am----> work done.
Explanation:
The explanation to this question can be given as:
Firstly we write the equation that is given in the question like, pressure, volume, gas constant, temperature. Then we covert the pressure and temperature into there possible values. Then We apply the formula of Ideal gas that is M=PV/RT1. where M is the mass, P is the pressure
, V is the volume
, R is the gas constant (0.08206 L·atm·K−1·mol−1) and T1 is the initial temperature. Then we use the second formula that is DQ=DU+W. Where DQ= Heat transfer, DU=MCV(T2-T1)and W= work. It is used for finding work.
