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Container A holds 717 mL of an ideal gas at 2.20 atm. Container B holds 124 mL of a different ideal gas at 4.20 atm. Container A and container B are glass spheres connnected by a tube with a stopcock. Container A is larger than container B. If the gases are allowed to mix together, what is the resulting pressure?

Respuesta :

Answer:

2.4949 atm

Explanation:

For Container A :  

Pressure = 2.20 atm

Temperature = T K

Volume = 717 mL = 0.717 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

2.20 atm × 0.717 L = n × 0.0821 L.atm/K.mol × T K  

⇒n = 19.2132 / T moles

For Container B :  

Pressure = 4.20 atm

Temperature = T

Volume = 124 mL = 0.124 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

Applying the equation as:

4.20 atm × 0.124 L = n × 0.0821 L.atm/K.mol × 298 K

⇒n = 6.3435 / T moles

Total moles = 19.2132 / T moles + 6.3435 / T moles = 25.5567 /T moles

Total Volume = 717 + 124 mL  =  841 mL = 0.841 L

Temperature = T

Using ideal gas equation as:

PV=nRT

Applying the equation as:

P × 0.841 L = 25.5567 /T × 0.0821 L.atm/K.mol × T

⇒Total P = 2.4949 atm

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