Our data are,
State 1:
![P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K](https://tex.z-dn.net/?f=P_1%3D%2010psi%3D68.95kPa%5C%5CV_1%20%3D%201ft%5E3%3D0.02831m%5E3%5C%5CT_1%20%3D%20100%5C%C2%B0F%20%3D%20310.93K)
State 2:
![P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3](https://tex.z-dn.net/?f=P_2%20%3D5psi%3D34.474kPa%5C%5CV_2%20%3D%203ft%5E3%3D0.0899m%5E3)
We know as well that ![3BTU=3.16kJ/K](https://tex.z-dn.net/?f=3BTU%3D3.16kJ%2FK)
To find the mass we apply the ideal gas formula, which is given by
![P_1V_1=mRT_1](https://tex.z-dn.net/?f=P_1V_1%3DmRT_1)
Re-arrange for m,
![m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7BP_1V_1%7D%7BRT_1%7D%5C%5Cm%3D%20%5Cfrac%7B68.95%2A0.02831%7D%7B%280.287%29310.9%7D%5C%5Cm%3D0.021893kg%3D0.04806lbm%5C%5C)
Because of the pressure, temperature and volume ratio of state 1 and 2, we have to
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
Replacing,
![T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F](https://tex.z-dn.net/?f=T_2%20%3D%20%5Cfrac%7BP_2V_2%7D%7BP_1V_1%7DT_1%5C%5CT_2%20%3D%5Cfrac%7B34.474%2A0.0844%7D%7B68.95%2A0.02831%7D%2A310.93%5C%5CT_2%20%3D%20464.217K%3D375.5%5C%C2%B0F)
For conservative energy we have, (Cv = 0.718)
![W = m C_v = 0.718 \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU](https://tex.z-dn.net/?f=W%20%3D%20m%20C_v%20%3D%200.718%20%20%5CDelta%20T%20%2Bdw%5C%5Cdw%20%3D%20W%20-%20mv%5CDelta%20T%5C%5Cdw%20%3D%203.16-%280.0218%2A0.718%29%28454.127-310.93%29%5C%5Cdw%20%3D%200.765kJ%3D0.72BTU)
Answer:i One way to solve the quadratic equation x2 = 9 is to subtract 9 from both sides to get one side equal to 0: x2 – 9 = 0. The expression on the left can be factored:
Explanation:
the key result indicate that how you will achieve the your goal