Answer to Question #235288 in Molecular Physics | Thermodynamics for Jimmy

Question

For a certain ideal gas R = 25.8 ft-lb_f/lb_m-⁰R and k = 1.09 (a) What are the values of c_p and c_v? (b) What mass of this gas would occupy a volume of 15 cu ft at 75 psia and 80⁰F? (c) If 30   Btu are transferred to this gas at constant volume in (b), what are the resulting temperature and pressure?

a. C_p = ____ Answer in 4 decimal places

C_v = _____ Answer in 4 decimal places

b. m = ______ lb Answer in 2 decimal places

c. T₂ = _____

P₂ = ______ psia

Answer 1

$R= 25.8\; \frac{ft \cdot lb}{lb \cdot R} \\ k=1.09 = \frac{c_p}{c_v} \\ v= 15 \;cu \;ft \\ p=75 \;psia$

a. For one mole of ideal gas we know that

$c_p -c_v =R \\ \frac{c_p}{c_v} -1 = \frac{R}{c_v} \\ k -1= \frac{R}{c_v} \\ c_v = \frac{R}{k-1} \\ c_v = \frac{25.8}{1.09-1} = 286.6667 \; BTU/F/lb \\ c_p =c_vk \\ c_p = 286.667 \times 1.09 = 312.4667 \;BTU/F/lb$

b. For ideal gas we know

$pV=mRT \\ m =\frac{pV}{RT} \\ m = \frac{75 \;lb \times 144\;in^2}{in^2 \times ft^2} \times \frac{15 \;ft^3 \times lb \;R}{25.8 ft \;bl \times (80 + 459.67)R} \\ m = \frac{75 \times 144 \times 15}{25.8 \times 539.67} \;lb \\ m = 11.635 \;lb$

c. Q = 30 BTU

$T_1 = 80 °F \\ Q=mc_vΔT \\ Q= mc_v(T_2 -T_1) \\ \frac{Q}{mc_v} +T_1 =T_2 \\ T_2 = \frac{30}{286.67 \times 11.635} +80 \\ T_2 = 80.009 \;°F$

For ideal gas we know

$pV=RT$

at constant volume we know

$\frac{p}{T} = constant \\ \frac{p_1}{T_1} = \frac{p_2}{T_2} \\ p_2 = \frac{p_1}{T_1} \times T_2 \\ p_2 = \frac{75 \times 80.009}{80} \\ p_2 = 75.0084 \;psia$