Suppose 200 ft^3 of air measured at 15 psia and 90 f is withdrawn from a 50- ft^3 tank containing air initially at 100 psia and 140 f. what is the pressure of the remaining air in the tank if its temperature is 130f

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

Explanation :

First we have to calculate the moles of air.

where,

P = pressure of gas =

conversion used : (1 psia = 0.068046 atm)

V = volume of gas =

conversion used :

T = temperature of gas =

= 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:

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

where,

P = pressure of gas =

conversion used : (1 psia = 0.068046 atm)

V = volume of gas =

conversion used :

T = temperature of gas =

= 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:

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

Now we have to determine the pressure of the remaining air in the tank if its temperature is .

where,

P = pressure of gas = ?

V = volume of gas =

conversion used :

T = temperature of gas =

= 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:

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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

would be false hope this out