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0.5 gm of nitrogen gas is contained in sealed vessel of 435 ml capacity at 25°c . Calculate the pressure exerted by the gas (R=0.0821 atm lit k-¹ mol-¹) .
Ans :- 725.04 mm hg
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Given info : 0.5 gm of nitrogen gas is contained in sealed vessel of 435 ml capacity at 25°C.
To find : the pressure exerted by the gas.
Solution : no of moles of nitrogen gas, n = given weight/molecular weight of N2
= 0.5/28 = 0.017 mol
volume of gas ( i.e., volume of vessel) , V = 435 ml = 0.435 Litre [ you know, 1l = 1000 ml]
temperature , T = 25°C = 273 + 25 = 298K
Now using formula, PV = nRT
⇒P = nRT/V
= (0.017 × 0.0821 × 298)/(0.435)
= 0.956
Now we know, 1 atm = 760 mmHg
So, 0.8267 atm = 0.956 × 760 mmHg = 726.56 mmHg (nearest to given answer this has happened because we take average value of molecular mass and others.)
Therefore the pressure exerted by the has is 725.04 mm Hg
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