Question

A 265-gallon storage tank contains 88.5 kg methane (ch4). if the temperature reaches 42 °c, what is the pressure inside the tank?

Answer 1

Given:

Volume of the tank (V) = 265 gallons

Mass of methane = 88.5 kg

Temperature T = 42 C

To determine:

The pressure P in the tank

Step-by-step explanation:

From ideal gas law:

PV = nRT ----(1)

where P = pressure, V = volume, n = moles, R = gas constant = 0.0821 L.atm/mol-K and T = temperature

Based on the given data

V = 265 gal = 1003.13 L

T = 42 + 273 = 315 K

moles of CH4 (n) = mass of CH4/molar mass = 88.5/16 = 5.531 moles

Based on eq (1) we have:

P = nRT/V = 5.531 * 0.0821*315/1003.13 = 0.143 atm

Ans: The tank pressure is 0.143 atm

Answer 2

Answer : The pressure inside the tank is 142.6 atm

Step-by-step explanation:

Using ideal gas equation:

`PV=nRT\\\\PV=(w)/(M)RT`

where,

P = pressure of gas = ?

V = volume of gas =
`265\text{ gallon}=1003.025L`

conversion used :
`(1\text{ gallon}=3.785L)`

T = temperature of gas =
`42^oC=273+42=315K`

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

w = mass of methane gas = 88.5 kg = 88500 g

M = molar mass of methane = 16 g/mole

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

`P* (1003.025L)=(88500g)/(16g/mole)* (0.0821L.atm/mole.K)* (315K)`

`P=142.6atm`

Therefore, the pressure inside the tank is 142.6 atm