Answer:
The inside Pressure of the tank is [tex]4499.12 lb/ft^{2}[/tex]
Solution:
As per the question:
Volume of tank, [tex]V = 0.25 ft^{3}[/tex]
The capacity of tank, [tex]V' = 50ft^{3}[/tex]
Temperature, T' = [tex]80^{\circ}F[/tex] = 299.8 K
Temperature, T = [tex]59^{\circ}F[/tex] = 288.2 K
Now, from the eqn:
PV = nRT (1)
Volume of the gas in the container is constant.
V = V'
Similarly,
P'V' = n'RT' (2)
Also,
The amount of gas is double of the first case in the cylinder then:
n' = 2n
[tex]\]frac{n'}{n} = 2[/tex]
where
n and n' are the no. of moles
Now, from eqn (1) and (2):
[tex]\frac{PV}{P'V'} = \frac{nRT}{n'RT'}[/tex]
[tex]P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}[/tex]
