Question

What volume is occupied by a `3.0*mol` volume of gas that is heated in piston to a temperature of `323.1*K` under a pressure of `0.12*atm`?

Answer 1

I presume you quote pressure with units of `"Pascal"`...`101.3*kPa-=1*atm`

Explanation:

And so `"pressure"=(12,156*Pa)/(101.3xx10^3*Pa*atm^-1)=0.12*atm`

And we solve the old Ideal Gas Equation....

`V=(nRT)/P=(3.0*molxx0.0821*(L*atm)/(K*mol)xx323.1*K)/(0.12*atm)`

...I make this volume over `600*L`.......