Respuesta :
Answer : The amount of air added to the tank will be, 1.2062 Kg.
Explanation :
First we have to calculate the volume of air by using ideal gas equation.
[tex]P_1V_1=\frac{m_1RT_1}{M}[/tex]
where,
[tex]P_1[/tex] = initial pressure of air = [tex]20psia=1.36atm[/tex]
conversion used : [tex]1psia=0.068046atm[/tex]
[tex]T_1[/tex] = initial temperature of air = [tex]70^oF=294.261K[/tex]
conversion used : [tex](70^oF-32)\frac{5}{9}+273.15=294.261K[/tex]
[tex]V_1[/tex] = initial volume of air = ?
[tex]m_1[/tex] = initial mass of air = [tex]20Ibm=9071.85g[/tex]
conversion used : [tex]1lbm=453.592g[/tex]
R = gas constant = 0.0821 L.atm/mole.K
M = molar mass of air
Now put all the given values in the above expression, we get:
[tex](1.36atm)\times V_1=\frac{(9071.85g)\times (0.0821L.atm/mole.K)\times (294.261K)}{M}[/tex]
[tex]V_1=\frac{161150.9299}{M}L[/tex]
Now we have to calculate the final amount of air by using ideal gas equation.
[tex]P_2V_2=\frac{m_2RT_2}{M}[/tex]
where,
[tex]P_2[/tex] = final pressure of air = [tex]23.5psia=1.599atm[/tex]
[tex]T_2[/tex] = final temperature of air = [tex]90^oF=305.37K[/tex]
[tex]V_2[/tex] = final volume of air = [tex]V_1=\frac{161150.9299}{M}L[/tex]
[tex]m_2[/tex] = final mass of air = ?
R = gas constant = 0.0821 L.atm/mole.K
M = molar mass of air
Now put all the given values in the above expression, we get:
[tex](1.599atm)\times (\frac{161150.9299}{M}L)=\frac{m_2\times (0.0821L.atm/mole.K)\times (305.37K)}{M}[/tex]
[tex]m_2=10278.074g[/tex]
Now we have to calculate the amount of air added to the tank.
[tex]m_2-m_1=10278.074g-9071.85g=1206.224g=1.2062Kg[/tex]
conversion used : (1 Kg = 1000 g)
Hence, the amount of air added to the tank will be, 1.2062 Kg.
The relation between the volume, the pressure, and the temperature is PV = mRT. Then the amount of air added to the tank is 1.2062 kg.
What is thermodynamics?
It is a branch of science that deals with heat and work transfer.
A rigid tank contains 20 lbm of air at 20 psi and 70°F. More air is added to the tank until the pressure and temperature rise to 23.5 psi and 90°F, respectively.
The ideal gas equation is
[tex]PV = \dfrac{mRT}{M}[/tex]
[tex]P_1 = 20 \ psi = 1.36 \ atm\\\\T_1 = 70 ^oF = 294.261\ K\\\\V_1 = \ \ ? \\\\m_1 = 20 \ lbm = 9071.85 \ kg[/tex]
R = 0.0821 L atm/mole K
M = molar mass of air
Now put all the given values in the ideal gas equation, we have
[tex]\rm V_1 = \dfrac{9071.85*0.0821*294.261}{1.36M}\\\\\\V_1 = \dfrac{161150.8299}{M}[/tex]
Now we have to calculate the final amount of air by using an ideal gas equation, we have
[tex]P_2 = 23.5 \ psi = 1.599\ atm\\\\T_2 = 90^oF = 305.37\ K\\\\V_2 = V_1\\\\m_2 = \ \ ?[/tex]
Then we have
[tex]m_2 = \dfrac{1.599M* \dfrac{161150.9299}{M}}{0.0821*305.37}\\\\m_2 = 10278.074 \ g[/tex]
Now we have to calculate the amount of air added to the tank will be
⇒ m₂ - m₁ = 10278.074 - 9071.85 = 1206.244 g = 1.12062 kg
Hence, the amount of air added to the tank is 1.2062 kg.
More about the thermodynamics link is given below.
https://brainly.com/question/7206767
