Question

Enter your answer in the provided box. After 0.600 L of Ar at 1.36 atm and 210°C is mixed with 0.200 L of O2 at 338 torr and 123°C in a 400.-mL flask at 27°C, what is the pressure in the flask? atm

Answer 1

Answer:

Here we have to Ideal gas equation.

PV=nRT where P=pressure, V=volume, n=number of moles, R=gas constant=0.0821 L atm mol^-1 K^-1, T=temperature.

Firstwe have to find out moles of Ar and O2 gases.

Given for Ar,

P=1.36 atm, V=0.6 L, T=210°C=210+273=483 K,

Therefore number of moles of Ar are

n(Ar)=PV/RT=(1.36 atm x 0.6 L)/(0.0821 L atm mol^-1 K^-1 x 483 K)

n(Ar)=0.02058 mol.

For O2, P=338 torr=0.4447 atm (since 1 atm=760 torr)

V=0.2 L, T=123°C=123+273=396 K,

Therefore number of moles of O2 are

n(O2)=(0.4447 atm x 0.2 L)/(0.0821 L atm mol^-1 K^-1 x 396 K)=0.00273 mol.

Therefore total number of moles,

n(total)=n(Ar) + n(O2)=0.02058 + 0.00273 mol=0.0233 mol.

And now volume=400 mL=0.4 L, T=27°C=27+273=300 K, P=?

P=nRT/V

P=(0.0233 mol x 0.0821 L atm mol^-1 K^-1 x 300 K)/(0.4 L)

P=1.4357 atm.

Therefore the pressure in the flask ~ 1.44 atm.

Please let me know if you have any doubt. Thanks.