CHEMISTRY COMMUNITY

Posted: **Thu Oct 13, 2011 11:02 pm**

Hello Professor Lavelle!
How did the book get 101.325 $\frac{J}{L\: atm}$ ?
Thank you! Katherine

Posted: **Thu Oct 13, 2011 11:02 pm**

101.325 is the conversion of kPa to atm (101.325 kPa = 1 atm). The units of pascals and liters are required for the work equation to have the resulting answer be in units of joules.
W (J) = -P (Pa) x [ $V_{2}-V_{1}$ ] (L)

Further, 1 J = 1 Pa*m³. To get the relationship between Joules and L*atm:
1 L*atm= $\frac{101325 Pa}{1 atm}$ x $\frac{10^{-3} m^3}{1 L}$ = 101.325 Pa*m³ = 101.325 J

CHEMISTRY COMMUNITY

Conversion Factor 101.325 J/L*atm

Conversion Factor 101.325 J/L*atm

Re: Conversion Factor 101.325 J/L*atm