Page 19 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:35 9in x 6in b985-ch01 Elementary Physical Chemistry

4 Elementary Physical Chemistry

5
3
c) In Pascal L K −1 mol −1 [1 atm = 1.01325 × 10 Pa; 1 L= 10 −3 m ],
5
3
R =1.01325 × 10 Pa × 22.414 × 10 −3 m mol −1 /273.16 K
3 −1 −1
=8.314 Pa m K mol
−1 −1
=8.314 k Pa L K mol (1.6)
d) In J K −1 mol −1 ,

2 −2 −1 −1 −1 −2
R =8.314 kg m s =8.314 J K mol [1 Pa = 1 kg m s ] (1.7)
3
6
2
e) In cgs units (V in cm , P in dyne/cm ,1 atm = 1.013×10 dyne cm −2 ),
6
3
R =(1.013 × 10 dyne cm −2 ) × (22, 414 cm mol −1 )/273.16 K
7
Also 1 erg = 10 J,
7
R =8.314 × 10 erg K −1 mol −1 (1.8)
f) In cal K −1 mol −1 (1 cal = 4.184 J),

−1 −1
R =1.987 calK mol (1.9)

Example 1.1. 50.0g of N 2 (M =28.0 g) occupies a volume of 750 mL at
298.15 K. Assuming the gas behaves ideally, calculate the pressure of the
gas in kPa.

Solution
P = nRT/V
−1 −1 −1
=(50.0g/28.0gmol ) × (0.0826 atm L K mol × 298.15 K)/0.750 L
3
=58.25 atm = 58.25 × 101.325 kPa/atm = 5.90 × 10 kP

1.6. Mixtures of Gases
The partial pressure of a gas in a mixture is defined as the pressure the
gas would exert if it alone occupied the whole volume of the mixture at the
same temperature. Dalton’s Law states that the total pressure is equal to