Question

A sealed 26-m3 tank is filled with 6000 moles of oxygen gas (O2) at an initial temperature of 270 K. The gas is heated to a final temperature of 440 K. The ATOMIC mass of oxygen is 16.0 g/mol, and the ideal gas constant is R = 8.314 J/mol·K = 0.0821 L·atm/mol·K. The final pressure of the gas is closest to
A) 0.31
B) 0.34
C) 0.33
D) 0.36
E) 0.29

Answer 1

Answer:

The final pressure of oxygen gas is 8.33 atm.

Explanation:

From the given data

V=26 m^3 or 26000 L

T1=270K

T2=440K

n1=6000 moles

R=0.0821 L.atm/molK

Now from the ideal gas equation

$P_2V_2=nRT_2\\P_2=\frac{nRT_2}{V_2}\\P_2=\frac{6000*0.0821*440}{26000}\\\\P_2=8.33 atm$

As the options given are not defined in which unit thus the final pressure of oxygen gas is 8.33 atm.

*The options are provided for a different question where

Answer 2

The final pressure of the gas in the sealed container is closest to 8.34 atm

How to determine the initial pressure

• Volume (V) = 26 m³ = 26 × 1000 = 26000 L
• Temperature (T) = 270 K
• Gas constant (R) = 0.0821 atm.L/Kmol
• Number of mole (n) = 6000 moles

• Initial Pressure (P) =?

The initial pressure of the gas can be obtained by using the ideal gas equation as follow:

PV = nRT

Divide both side by V

P = nRT / V

P = (6000 × 0.0821 × 270) / 26000

P = 5.12 atm

How to determine the final pressure  

• Temperature (T₁) = 270 K

• Initial pressure (P₁) = 5.12 atm
• Volume = constant
• Final temperature (T₂) = 440 K

• Final pressure (P₂) =?

The final pressure of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

Since the volume is constant, we have:

P₁ / T₁ = P₂ / T₂

5.12 / 270 = P₂ / 440

Cross multiply

270 × P₂ = 5.12 × 440

Divide both side by 270

P₂ = (5.12 × 440) / 270

P₂ = 8.34 atm

Learn more about gas laws:

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