Question

how many moles of gas are in 30.0 l of a contained tube with a temperature of 300.k and pressure at 200. atm? use r

Answer 1

234.61 mol

Step-by-step explanation:

According to the Ideal Gas Law:

PV=nRT

P= 200 atm

V= 30.0 L

T= 300 K

R= 0.0821
`((atm)(L))/((mol)(K))`

`n=(PV)/(RT)`

`n=((200atm)(30.0L))/(0.0821((atm)(L))/((mol)(K))(300K))\\n=234.61 mol`