Asample of 0.020 mole of oxygen gas is confined at 227°c and 0.50 atmospheres. what would be the pressure of this sample at 27°c and the same volume?

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0.3 atm

Explanation:-

Number of moles n = 0.020 mol

Pressure P 1 = 0.5 atm

Temperature T 1 = 227 C + 273 = 500 K

We know the value of universal Gas constant R = 0.082 L atm K−1 mol−1

We use the relation PV = nRT

V = nRT /P

Plugging in the values

V = 0.020 mol x (0.082 L atm K−1 mol−1 ) x 500 K / (0.5 atm)

= 1.64 L

Now the new temperature T 2 = 27 C + 273 = 300 K.

Volume is same = 1.64 L

Number of moles n is same = 0.020 mol

Once again we use the relation

PV = nRT

New Pressure P = nRT / V

= 0.020 x 0.082 L atm K−1 mol−1 x 300 K / 1.64 L

= 0.3 atm

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