What is the final pressure of argon gas after isochoric heating?

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SUMMARY

The final pressure of 0.54 mol of argon gas in a 90 cm³ container after isochoric heating to 400 °C is calculated to be 33,571.932 kPa using the ideal gas law equation PV = nRT. The calculation utilizes the gas constant R = 8.314 m³Pa/K mol and the absolute temperature of 673 K. This pressure value is significantly high due to the small volume of the container, which is only 90 mL, leading to a high density of gas molecules.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of temperature conversion to Kelvin
  • Familiarity with units of pressure (kPa, atm, psi)
  • Basic concepts of isochoric processes in thermodynamics
NEXT STEPS
  • Learn about the implications of gas behavior under isochoric conditions
  • Explore the concept of pressure-volume relationships in ideal gases
  • Study the conversion between different pressure units (kPa to atm, psi)
  • Investigate real gas behavior and deviations from the ideal gas law
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Students studying thermodynamics, physics enthusiasts, and anyone involved in gas law applications in chemistry or engineering.

kristibella
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Homework Statement


0.54 mol of argon gas is admitted to an evacuated 90 cm3 container at 10 o C. The gas then undergoes an isochoric heating to a temperature of 400 oC.

What is the final pressure of the gas (in kPa)?

Homework Equations


PV = nRT

The Attempt at a Solution


P(9.0x10-5 m3) = (.54 mol)(8.314 m3Pa/K mol)(673 K)
P = 33,571,932 Pa = 33,571.932 kPa

This number just seems way too large to me, which is why I'm questioning it. Thanks for any help you can provide!
 
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Convert it to atmospheres or psi if you want...
 
That's a correct answer. 90 mL for half a mole of gas is a pretty small volume even at room temperature (remember at STP it should occupy about 12L).
 

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