Ideal Gas Pressure: Solve Physics HW

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SUMMARY

The discussion focuses on calculating the change in pressure of one mole of an ideal gas when the temperature increases from 0°C to 61°C at a constant volume of 94 liters. Using the ideal gas law (PV=nRT), the initial pressure (P1) is calculated as 0.2386 atm and the final pressure (P2) as 0.2918 atm, resulting in a change in pressure of 0.0533 atm. The importance of significant digits in reporting the final answer is also highlighted, emphasizing that precision should align with the given data.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Basic knowledge of temperature conversion (Celsius to Kelvin)
  • Familiarity with significant figures in scientific calculations
  • Ability to perform unit conversions (atm, liters, moles)
NEXT STEPS
  • Review the Ideal Gas Law applications in various conditions
  • Learn about significant figures and their importance in scientific reporting
  • Explore temperature conversions and their impact on gas behavior
  • Investigate real gas behavior versus ideal gas assumptions
USEFUL FOR

Students studying physics, particularly those tackling thermodynamics and gas laws, as well as educators looking for examples of ideal gas calculations and significant figure applications.

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Homework Statement



This is for my physics homework and I haven't done gen chem in a while so I was wondering if this was right:

One mole of an ideal gas at 0◦C is held at a constant volume of 94 liter. Find the change in pressure if the temperature increases by 61◦C. Answer in units of atm.

Homework Equations



PV=nRT

The Attempt at a Solution



#1:
P1=?
V= 94 L
n= 1 mole
R= 0.0821 L atm / mol K
T1= 273.15 K

#2:
P2= ?
V= 94 L
n= 1 mole
R= 0.0821 L atm / mol K
T2= 334.15 K

Work:

#1:
P1 = \frac{nRT}{V} = \frac{0.0821 x 1 x 273.15}{94} = 0.238570372 atm

#2:
P2 = \frac{nRT}{V} = \frac{0.0821 x 1 x 334.15}{94} = 0.291848301 atm

change in pressure = P2-P1 = 0.291848301 - 0.238570372 = 0.053277659 atm

Is that right? TIA
 
Last edited:
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In general yes, although you abuse significant digits. They are not treated seriously in physics (after all, they are just a poor relative of the real accuracy), but listing nine digits when you are given only two for value doesn't make sense.
 
Borek said:
In general yes, although you abuse significant digits. They are not treated seriously in physics (after all, they are just a poor relative of the real accuracy), but listing nine digits when you are given only two for value doesn't make sense.

Thank you, oh I know the digits are long, I plan to answer it as: .0533 atm
 

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