If we have 16 gm of Oxygen gas at 273 C and 1 atm pressure, what would be its volume?
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PV=nRT
where,P=1atm. V=?
n=16/16=1mole
R=1/12 and T=273+273=546K
therefore,1×V=1×1/12×546
Answered by
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Look....we know the ideal gas equation as
pv=nRT, where p is the pressure(in atm), v= volume(in litre) , n=no. of moles, R=universal gas constant and T=Temperature{in kelvin}...
so P = 1atm according to question and v we have to find so
v=n/p R ×T
= 16/16×2 × 0.082 × 546
= 22.386 litre...
I hope it helps...
pv=nRT, where p is the pressure(in atm), v= volume(in litre) , n=no. of moles, R=universal gas constant and T=Temperature{in kelvin}...
so P = 1atm according to question and v we have to find so
v=n/p R ×T
= 16/16×2 × 0.082 × 546
= 22.386 litre...
I hope it helps...
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