If 500 ml of nitrogen is collected over water at 0.342 bar and 27 °c, then what will be the volume of dry gas at stp? it is given that aqueous tension at 27 °c is 0.0218 bar
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We know the relation, P1V1/T1=P2V2/T2
We have P1= 0.342-0.0218 = 0.3202 (aqueous tension has to be subtracted)
V1= 500 ml
T1 = 300 K
P2= 1 atm pressure at STP
V2 = ?
T2 = 273K
Substituting values in equation we get :
0.3202×500/300=1×V2/273
0.5336 ×273 =V2
V2 = 145.7mL
Hope it's helpful to uh..
We have P1= 0.342-0.0218 = 0.3202 (aqueous tension has to be subtracted)
V1= 500 ml
T1 = 300 K
P2= 1 atm pressure at STP
V2 = ?
T2 = 273K
Substituting values in equation we get :
0.3202×500/300=1×V2/273
0.5336 ×273 =V2
V2 = 145.7mL
Hope it's helpful to uh..
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