Suppose 200 ft^3 of air measured at 15 psia and 90 f is withdrawn from a 50- ft^3 tank containing air initially at 100 psia and 140 F. What is the pressure of the remaining air in the tank if its temperature is 130F

Question

23-08-2019
Physics

contestada

Suppose 200 ft^3 of air measured at 15 psia and 90 f is withdrawn from a 50- ft^3 tank containing air initially at 100 psia and 140 F. What is the pressure of the remaining air in the tank if its temperature is 130F

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Alleei Alleei · Answer 1 · 2019-09-02T13:04:00+02:00

Answer : The pressure of the remaining air in the tank is 115.56 atm.

Explanation :

First we have to calculate the moles of air.

[tex]PV=n_1RT[/tex]

where,

P = pressure of gas = [tex]15psia=1.02069atm[/tex]

conversion used : (1 psia = 0.068046 atm)

V = volume of gas = [tex]200ft^3=5663.37L[/tex]

conversion used : [tex]1ft^3=28.3168L[/tex]

T = temperature of gas = [tex]K=(90^oF-32)\times \frac{5}{9}+273=305.372K[/tex]

[tex]n_1[/tex] = number of moles of air = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

[tex](1.02069atm)\times (5663.37L)=n_1\times (0.0821L.atm/mol.K)\times (305.372K)[/tex]

[tex]n_1=230.57mole[/tex]

Now we have to calculate the moles of air in tank initially.

[tex]PV=n_2RT[/tex]

where,

P = pressure of gas = [tex]100psia=6.8046atm[/tex]

conversion used : (1 psia = 0.068046 atm)

V = volume of gas = [tex]50ft^3=1415.84L[/tex]

conversion used : [tex]1ft^3=28.3168L[/tex]

T = temperature of gas = [tex]K=(140^oF-32)\times \frac{5}{9}+273=333.15K[/tex]

[tex]n_2[/tex] = number of moles of air = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

[tex](6.8046atm)\times (1415.84L)=n_2\times (0.0821L.atm/mol.K)\times (333.15K)[/tex]

[tex]n_2=352.24mole[/tex]

Now we have to calculate the remaining moles in the tank.

[tex]n=n_2-n_1=352.24mole-230.57mole=121.67mole[/tex]

Now we have to determine the pressure of the remaining air in the tank if its temperature is [tex]130^oF[/tex].

[tex]PV=n_2RT[/tex]

where,

P = pressure of gas = ?

V = volume of gas = [tex]50ft^3=1415.84L[/tex]

conversion used : [tex]1ft^3=28.3168L[/tex]

T = temperature of gas = [tex]K=(130^oF-32)\times \frac{5}{9}+273=327.594K[/tex]

[tex]n_2[/tex] = number of moles of air = 121.67 mole

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

[tex]P\times (28.3168L)=(121.67mole)\times (0.0821L.atm/mol.K)\times (327.594K)[/tex]

[tex]P=115.56atm[/tex]

Therefore, the pressure of the remaining air in the tank is 115.56 atm.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-08T10:59:29+02:00

The pressure of the remaining air in the tank if its temperature is 130Fis mathematically given as

P=115.56atm

What is the pressure of the remaining air in the tank if its temperature is 130F?

Question Parameter(s):

200 ft^3 of air measured

at 15 psia and 90 f is withdrawn from

a 50- ft^3 tank containing air initially at 100 psia and 140 F.

Generally, the equation for the ideal gas is mathematically given as

PV=nRT

Therefore

(1.02069atm) * (5663.37)=n1 *(0.0821)* (305.372)

n1=230.57mole

In conclusion, for n2 number of moles

(6.8046)*(1415.84L)=n2*(0.0821)* (333.15K)

n_2=352.24mole

Hence Remaining moles

nr=n_2-n_1

nr=352.24mole-230.57mole

nr=121.67mole

Hence, the Pressure of air in tank

P*(28.3168L)=(121.67*(0.0821)* (327.594K)

P=115.56atm