Question

A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure inside the tank. One standard cubic foot of air occupies foot at standard 'temperature and pressure (T 59° F and p= 2116 lb/ft2). one cubic

Answer 1

Answer:

The inside Pressure of the tank is $4499.12 lb/ft^{2}$

Solution:

As per the question:

Volume of tank, $V = 0.25 ft^{3}$

The capacity of tank, $V' = 50ft^{3}$

Temperature, T' = $80^{\circ}F$ = 299.8 K

Temperature, T = $59^{\circ}F$ = 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

$\]frac{n'}{n} = 2$

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

$\frac{PV}{P'V'} = \frac{nRT}{n'RT'}$

$P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}$