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0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a volume of 400 cm^3 .What is the final pressure of the gas?

Respuesta :

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

[tex]PV=nRT[/tex]

[tex]P_{i}=\dfrac{nRT}{V}[/tex]

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

[tex]P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}[/tex]

[tex]P_{i}=5.75\times10^{6}\ Pa[/tex]

[tex]P_{i}=56.827\ atm[/tex]

We need to calculate the final temperature

Using relation pressure and volume

[tex]P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}[/tex]

[tex]P_{2}=\dfrac{56.827\times70}{400}[/tex]

[tex]P_{2}=9.94\ atm[/tex]

Hence, The final pressure of the gas is 9.94 atm.

Eduard22sly

The final pressure of the gas given that it undergoes an isothermal expansion is 9.95 atm

How to determine the initial pressure

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

  • Number of mole (n) = 0.16 mole
  • Volume (V) = 70 cm³ = 70 / 1000 = 0.07 L
  • Temperature (T) = 30 °C = 30 + 273 = 303 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Pressure (P) =?

P = nRT / V

P = (0.16 × 0.0821 × 303) / 0.07

P = 56.86 atm

How to determine the final pressure

  • Initial volume (V₁) = 70 cm³
  • Initial pressure (P₁) = 56.86 atm
  • Final Volume (V₂) = 400 cm³
  • Final pressure (P₂) =?

P₁V₁ = P₂V₂

56.86 × 70 = P₂ × 400

Divide both side by 400

P₂ = (56.86 × 70) / 400

P₂ = 9.95 atm

Learn more about gas laws:

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