Answer to Question #100124 in General Chemistry for Beverlie

Question

1. You have a cylinder of argon gas at 19.8 atm of pressure at 19 oC. The volume of the cylinder is 50.0 L. What would be the volume of the gas if you were to decrease the pressure to 0.974 atm? Assume constant temperature.
2. Helium gas at 22o C and 1.00 atm occupies a vessel has a volume of 2.54 L. What would be the volume if the gases temperature was decreased to -197 oC? Assume constant pressure
3. A cylinder contains 91.3 g of oxygen gas. If the volume of the cylinder is 8.58 L, what is the pressure of the gas in atmospheres if gas temperature is 21 oC? Calculate the pressure using the ideal gas equation and the van der Waals equation.

Answer 1

1.      P1=19.8 atm, Tc=292.15K, V1=0.05m3, P2= 0.974atm.

$P1V1=P2V2$

V2= (19.8*0.05)/0.974

V2=1.016m3

2.      T1=295.15K, Pc=1atm, V1=0.0254m3, T2=76.15K

$(V1/T1) = (V2/T2)$

V2= (76.15*0.0254)/295.15

V2=0.006553m3

3.      m= 91.3g, V= 0.00885, T= 294.15K, P=…

Using the ideal gas equation;$PV=nRT$

n=91.3/32=2.853moles. and R=8.314

P= (2.853*8.314*294.15)/0.00885

P=788383.22pa

P=7.78atm

Now using the van der waals equation; ${P+ a(n2/V 2}(V-nb)=nRT$

a=1.382, b=0.03186

p +1.382(8.14/0.000078) (0.00885-2.853*0.03186) =2.853*8.314*294.15

p+(14422410.26) (-0.08204658) =6977.191524.

p=1183309.437pa

p=11.7atm