Consider 8.2 L of an ideal gas of weight 9.0 gm and at 300 K temperature and 1 atm pressure. What is the molecular mass of gas?

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Solution

We know, $P V = \frac{W}{M} R T \Rightarrow M = \frac{9 \times 0.082 \times 300}{1 \times 8.2} = 27 g m o l^{- 1}$

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Consider 8.2 L of an ideal gas of weight 9.0 gm and at 300 K temperature and 1 atm pressure. What is the molecular mass of gas?

We know,PV=WMRT⇒M=9×0.082×3001×8.2=27 gmol−1

We know, $P V = \frac{W}{M} R T \Rightarrow M = \frac{9 \times 0.082 \times 300}{1 \times 8.2} = 27 g m o l^{- 1}$