calculate the temperature, in k, of 2.20 moles of gas occupying 3.30 l at 3.50 atm.



From The ideal gas equation pv=nrt
3.5 atm is equal to 3.5 × 101325 =354638 pascal
3.30 litre = 0.0033
therefore
pv=nrT
354638 × 0.0030 = 2.20 × 8.314 ×T
T =(354638 × 0.0030)/( 2.20 × 8.314)
T  =

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