Question

A 0.525 m^3 steel tank contains nitrogen at a gauge pressure of 485 kPa while sitting on a loading dock at a temperature of -14 Degrees Celsius. If the tank is set next to a furnace and the temperature of the tank and gas rise to 42.0 degrees Celsius. what is the new gauge pressure of the nitrogen in the tank? Assume that the tank is a rigid container and that the nitrogen acts as an ideal gas.

Answer 1

Explanation:

From the ideal gas law, we know that

`PV=\text{nrT}`

where

P = pressure

V = volume

n = number of moles of gas

r = gas constant

T = temperature.

Know in our case, we know that for a certain number of moles, if V = 0.525 m^3 , P = 485 kPa, then T = -14 celsius. Putting these values (by first converting them to metric units) gives

`(0.525m^3)(485\cdot10^3P)=nr(-14+273.15K)`

solving for nr gives

`nr=((0.525m^3)(485\cdot10^3P))/((-14+273.15)K)`

which evaluates to

`nr=982.54`

With the value of n*r in hand, we now find the pressure at 42 degrees celsius:

`P(0.525m^3)=(982.54)(42+273.15)K`

Solving for P ( the pressure) gives

`P=((42+273.15)(982.54))/(0.525)`
`P=589805\approx589.805\text{kpa}`

The new gauge pressure is 589.805 kPa.