Question

The volume of 1 gram mole of oxygen gas is 22400 ml at 760 mm. Calculate the temperature of the gas. (Given, R=0.082 lit-atmos deg ' molº!)​

Answer 1

Answer:

22400 ml. of oxygen at 273 K temperature and 3-atm pressure = 22400 ×3/1= 67200 ml at STP.

So 224 ml of oxygen at 273 K and 3-atm = 224 ×3= 672 ml at STP

The number of Oxygen molecules present in 22400 ml of gas at STP=N

(where N= Avagadro Number)

Therefore 224 ml of oxygen at 273K and 3-atm pressure will contain =N×672/22400 × N molecules.

=3/100×6.022×10

23

=18.066×10

21

Molecules =1.8066×10

22

molecules

Answer 2

Answer

✩ PV = nRTP = 760 mm Hg = 1atmV = 22400ml = 22.4 Lmoles = 1/16* T = PV /nR= 1x 22.4 x 16 / 0.0821= 358.4 / 0.0821= 4365.4 C= 4638 K