Answer to Question #295916 in Molecular Physics | Thermodynamics for dandan

Question

Three pounds of perfect gas with R=38 ft.lb/lb.R and k= 1.667 have 300 Btu of heat added during a reversible nonflow constant pressure change of state. The initial temperature is 100 °F. Determine the

(A) final temperature

(B) change in heat ∆H

(C) W

(D) ∆U

(E) ∆S

Answer 1

(1)

$C_p=\frac{kar}{k-1}=\frac{1.667\times38}{1.667-1}=\frac{94.97}{778}=0.1221BTU/lb R$

$Q=mc_p(T_2-T_1)$

$300=3\times0.1221(T_2-560)$

$T_2=1379=919°F$

(B)

$Q=mc_p(T_2-T_1)=H$

$H=300BTU$

(C)

$W=p(v_2-v_1)=p(\frac{mRT}{p_2}-\frac{mRT_1}{p_1})$

$W=pmR(\frac{T_2}{p}-\frac{T_1}{p})=mR(T_2-T_1)$

$W=3\times38\times(1379-560)=120.008BTU$

$(D)$

$C_v=\frac{R}{k-1}=\frac{38}{1.677-1}\times\frac{1}{778}=0.0732BTU/lb.R$

$∆U=mc_v(T_2-T_1)$

$∆U=3\times0.0732\times(1379-560)=179.85BTU$

(e)

$∆S=mc_pln\frac{T_2}{T_2}=4\times0.1221ln\frac{1379}{560}$

$∆S={0.3301}=BTU/R$