Answer to Question #237531 in Molecular Physics | Thermodynamics for Randal Rodriguez

Question

4. A fluid system undergoes a non-flow frictionless process from 4.5 ft 3 to a final volume 1.5 ft 3 in

accordance with the defining relation P= 60/V + 30 psi, where volume in ft 3 . During the process the

system reject 30 BTU of heat. Determine the change in enthalpy and internal energy.

Answer 1

Given:

$V_1 = 4.5\; ft^3 \\ V_2 = 1.5 \; ft^3 \\ Q_{1-2} = - 30 \; BTU \\ P = \frac{60}{V} + 30 \; psi$

we know that work done

$W_{1-2} = \int^{V_2}_{V_1} PdV = \int^{V_2}_{V_1} [\frac{60}{V} +30]dV \\ W_{1-2} = \int^{V_2}_{V_1} \frac{60}{V} dV + \int^{V_2}_{V_1} 30dV \\ W_{1-2} = 60[ln(V)]^{V_2}_{V_1} + 30[V]^{V_2}_{ V_1} \\ W_{1-2} = 60 ln(\frac{V_2}{V_1}) + 30(V_2-V_1) \\ W_{1-2} = 60ln(\frac{1.5}{4.5}) + 30(1.5 - 4.5) \\ W_{1-2} = -155.9167 \; psi \cdot ft^3$

we know that

$1 \; psi = 144 \; lbf/ft^2 \\ 1 \; BTU = 778 \; ft \cdot lbf \\ W_{1-2} = \frac{-155.9167 \times 144}{ 778} \; BTU \\ W_{1-2} = -28.86 \; BTU$

From first law of thermodynamics

$Q = ΔU + W \\ ΔU = Q - W \\ ΔU = -30 - (-28.86)$

ΔU = -1.14 BTU Change in inlernal energy

Now change in enthalpy is given as

$ΔH = ΔU + (P_2V_2 -P_1V_2) \\ P_1=\frac{60}{V_1} + 30 = \frac{60}{4.5} + 30 + 43.33 \; psi \\ P_1V_1=43.33 \times 4.5=195\; psi \cdot ft^3= \frac{195 \times 144}{778} = 36.09 \; BTU \\ P_2 = \frac{60}{V_2} + 30 = \frac{60}{1.5} + 30 = 70 \; psi \\ P_2V_2 = 70 \times 1.5 = 105 \; psi \cdot ft^3 = \frac{105 \times 144}{778} = 19.43 \; BTU ΔH = ΔU + (P_2V_2 – P_1V_1) \\ = - 1.14 + (19.43 -36.09) \\ ΔH = - 17.8 \;BTU$

change in enthalpy