Question

natural gas is stored in a spherical tank at a temperature of 10°c. at a given initial time, the pressure in the tank is 100 kpa gage, and the atmospheric pressure is 100 kpa absolute. sometime later, after considerably more gas is pumped into the tank, the pressure in the tank is 200 kpa gage, and the temperature is still 10°c. what will be the ratio of the mass of natural gas in the tank when p 200 kpa gage to that when the pressure was 100 kpa gage?

Answer 1

Step-by-step explanation:

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Answer 2

`(m_f)/(m_i)=1.5`

Step-by-step explanation:

We start by writting the ideal gas equation
`PV=nRT=(m)/(M)RT`, where M is the molar mass of the natural gas. Putting on the right side the variables that remain constant we have
`(P)/(m)=(RT)/(MV)`, so for our initial and final states we must have
`(P_i)/(m_i)=(P_f)/(m_f)`, which is the same as
`(m_f)/(m_i)=(P_f)/(P_i)`, which is what we want.

The pressure in the tank is
`P=P^(gage)+P_(atm)`, so for our values we have:

`(m_f)/(m_i)=(P_f)/(P_i)=(P^(gage)_f+P_(atm))/(P^(gage)_i+P_(atm))=(200kPa+100kPa)/(100kPa+100kPa)=1.5`