A 325-mL sample of air is at 720.0 torr and 30.0 °C. When the temperature is...

A 325-mL sample of air is at 720.0 torr and 30.0 °C. When the temperature is increased to 60.0 °C and the pressure is increased to 900.0 torr, the new volume of the gas will be _______ mL.

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Expert Solution

PV = nRT

P = 720 torr = 0.947 atm

T = 30 oC = 30 + 273 = 303 K

n = PV/RT

= 0.947 x 0.325/0.08205 x 303

= 0.0124 mols

When T = 60 oC = 60 + 273 = 333 K

P = 900 torr = 1.184 atm

Volume = 0.0124 x 0.08205 x 333/1.184 = 0.286 L = 286 ml

So the final gas volume = 286 ml

queen_honey_blossom answered 1 year ago

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