Question

What is the temperature of 3.3 moles of gas held at a pressure of 12 atm and in a container with a volume of 50 liters?

Answer 1

`2187.9K= T`

Explanation:

This problem requires the Ideal Gas Law equation since we are involving the mole value of the gas.

That equation is
`PV=nRT`
`P=` Pressure in atm
`V=` Volume in L
`n=` moles of gas
`R=0.0831(atm*L)/(mol*K)` - the gas law constant
`T=` Temperature in K

`P=12atm`
`V=50L`
`n=3.3` moles
`R=0.0831(atm*L)/(mol*K)` - the gas law constant
`T=?`

`12atm(50L)=3.3mol(0.0831(atm*L)/(mol*K))T`

`(12atm(50L))/(3.3mol(0.0831(atm*L)/(mol*K)))=T`

`(12cancel(atm)(50cancel(L)))/(3.3cancel(mol)(0.0831(cancel(atm)*cancel(L))/(cancel(mol)*K)))=T`

`(600/(.27423))K= T`

`2187.9K= T`