Question

3- A piston cylinder system contains 1.0 lbm of R134a at $P_1$ = 40 psi and $T_1$ = 80 °F. It expands isothermally to $P_2$ = 10 psi. Find: a) Change in the internal energy of R134a, $U_2 - U_1$. (2 points) b) Change of the entropy of R134a, $S_2 - S_1$. (2 points) c) The heat added to the R134a, $Q_{1-2}$. (3 points) d) The boundary work, $W_{1-2}$. (3 points) Note: °R = °F + 460

          3- A piston cylinder system contains 1.0 lbm of R134a at $P_1$ = 40 psi and
$T_1$ = 80 °F. It expands isothermally to $P_2$ = 10 psi. Find:
a) Change in the internal energy of R134a, $U_2 - U_1$. (2 points)
b) Change of the entropy of R134a, $S_2 - S_1$. (2 points)
c) The heat added to the R134a, $Q_{1-2}$. (3 points)
d) The boundary work, $W_{1-2}$. (3 points)
Note: °R = °F + 460

3- A piston cylinder system contains 1.0 lbm of R134a at P1 = 40 psi and
T1 = 80 °F. It expands isothermally to P2 = 10 psi. Find:
a) Change in the internal energy of R134a, U2 - U1. (2 points)
b) Change of the entropy of R134a, S2 - S1. (2 points)
c) The heat added to the R134a, Q1-2. (3 points)
d) The boundary work, W1-2. (3 points)
Note: °R = °F + 460

Added by Joshua D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/15/2023

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Therefore, the change in internal energy is zero. Show more…

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3- A piston cylinder system contains 1.0 Ibm of R134a at $P_1$ = 40 psi and $T_1$ = 80 °F. It expands isothermally to $P_2$ = 10 psi. Find: a) Change in the internal energy of R134a, $U_2$ - $U_1$. (2 points) b) Change of the entropy of R134a, $S_2$ - $S_1$. (2 points) c) The heat added to the R134a, $Q_{1-2}$. (3 points) d) The boundary work, $W_{1-2}$. (3 points) Note: °R = °F + 460