One litre of gas A at 2 atm pressure at 27 °C and two litreof gas B a

1.	One litre of gas A at 2 atm pressure at 27 °C and two litreof gas B at 3 atm pressure at 127 °C are mixed in a 4 litrevessel. The temperature of the mixture is maintained a327 °C. What is the total pressure of the gaseous mixture?
Answer» both gases are initially at the same absolute temperature T, Let Vo=the initial volume of oxygen=1 L=0.001 cubic meters The #moles of oxygen, No, is then by The Ideal Gas Law No=PoVo/(RT) Since Po=1 atm = 101,325 Pascals No=(101,325 Pa)(0.001 m^3)=101/(RT) Vn=the initial volume of nitrogen=0.002 m^3 Pn=the initial pressure of nitrogen=101,325/2=50,000 Pa Nn=the #moles of nitrogen=(PnVn)/(RT) =(50,000 Pa)(0.002 m^3)/(RT)=50/(RT) The combination of the two gases forms a new gas with N=No+Nn moles=150/(RT) The volume of that gas is given as V=1 L=0.001 m^3 The pressure is then P=NRT/0.001=(150/RT)(RT)/0.001 =150/0.001=150,000 Pa=1.5 atm This was probably not the fastest solution but it’s correct gjtwjue baueoeypwyosjhdg

One litre of gas A at 2 atm pressure at 27 °C and two litreof gas B at 3 atm pressure at 127 °C are mixed in a 4 litrevessel. The temperature of the mixture is maintained a327 °C. What is the total pressure of the gaseous mixture?

Answer»

both gases are initially at the same absolute temperature T,

Let Vo=the initial volume of oxygen=1 L=0.001 cubic meters

The #moles of oxygen, No, is then by The Ideal Gas Law

No=PoVo/(RT)

Since Po=1 atm = 101,325 Pascals

No=(101,325 Pa)(0.001 m^3)=101/(RT)

Vn=the initial volume of nitrogen=0.002 m^3

Pn=the initial pressure of nitrogen=101,325/2=50,000 Pa

Nn=the #moles of nitrogen=(PnVn)/(RT)

=(50,000 Pa)(0.002 m^3)/(RT)=50/(RT)

The combination of the two gases forms a new gas with

N=No+Nn moles=150/(RT)

The volume of that gas is given as V=1 L=0.001 m^3

The pressure is then

P=NRT/0.001=(150/RT)(RT)/0.001

=150/0.001=150,000 Pa=1.5 atm

This was probably not the fastest solution but it’s correct

gjtwjue baueoeypwyosjhdg