When 0 653 L of Ar at 1 20 atm and 227C is mixed with 0 278 Lof O2 at ...

pv nRT no of moles of argon n Ar PVRT 12065300821500 0019moles ....

When 0.653 L of Ar at 1.20 atm and 227Â°C is mixed with 0.278 L of...

4.2 Ratings (596 Votes)

pv = nRT

no of moles of argon n_Ar = PV/RT

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.2*0.653/0.0821*500

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.019moles

no of moles of O2 n_O2Â Â Â = PV/RT

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â p Â Â Â = 501/760 = 0.659atm

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â TÂ Â Â Â Â Â = 127 +273 = 400K

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â n_O2Â Â Â = 0.659*0.278/0.0821*400 = 0.00557moles

total no of moles nAr + n_{O2Â Â Â Â} = 0.019 +0.00557 = 0.02457moles

Â Â Â Â Â PV = nRT

Â Â Â Â Â Â PÂ Â = nRT/V

Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.024457*0.0821*300/0.4Â Â = 1.5atm >>>> anser

pressure in the flask = 1.5atm

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