Why multiply atm by atmospheric pressure in pascals?

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SUMMARY

The discussion focuses on the calculation of work done by an ideal gas during a two-step process involving pressure and volume changes. Specifically, the pressure drops from 2.7 atm to 1.4 atm, and the gas expands from a volume of 6.8 L to 13.1 L. The work done is calculated using the formula W = P(Vc - Vb), where the pressure must be converted from atm to pascals (1.4 atm multiplied by 1.013e5 pascals). The final result of the work done is 893.466 J, confirming that pressure in pascals multiplied by volume in cubic meters yields energy in joules.

PREREQUISITES
  • Understanding of ideal gas laws
  • Knowledge of unit conversions (atm to pascals)
  • Familiarity with work-energy principles in thermodynamics
  • Basic understanding of volume measurements (liters to cubic meters)
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn about unit conversions, specifically between atmospheric pressure and pascals
  • Explore the relationship between pressure, volume, and work in thermodynamic processes
  • Investigate the significance of energy units in physics, particularly joules
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Students and professionals in physics, engineering, and thermodynamics who are looking to deepen their understanding of gas behavior and energy calculations in thermodynamic processes.

gibson101
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Consider the following two-step process. Heat is allowed to flowout of an ideal gas at constant volume so that its pressure dropsfrom PA = 2.7 atmto 1.4 atm. Then the gas expands at constant pressure, from avolume of 6.8 L to VC = 13.1 L where the temperature reaches its originalvalue. See Fig. 15-27.

In the following work, why is 1.4 atm multiplied by 1.013e5 pascals? And to clarify, PA=2.7 just means pressure A.
Figure 15-27
(a) Calculate the total work done by the gas in the process (in joules).
Work done by gas along the parts BC W2= PdV
= P( Vc-Vb)

= 1.4*1.013*105( 13.1-6.8)*10-3
=893.466J
 
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gibson101 said:
Consider the following two-step process. Heat is allowed to flowout of an ideal gas at constant volume so that its pressure dropsfrom PA = 2.7 atmto 1.4 atm. Then the gas expands at constant pressure, from avolume of 6.8 L to VC = 13.1 L where the temperature reaches its originalvalue. See Fig. 15-27.

In the following work, why is 1.4 atm multiplied by 1.013e5 pascals? And to clarify, PA=2.7 just means pressure A.
Figure 15-27
(a) Calculate the total work done by the gas in the process (in joules).
Work done by gas along the parts BC W2= PdV
= P( Vc-Vb)

= 1.4*1.013*105( 13.1-6.8)*10-3
=893.466J

It's a unit conversion. The equation requires that the pressure be in pascals, and 1 atm = 1.01325 x 105 pascals.
 
So to get to joules of work, you have to multiply pascals times volume? And why is the answer divided by a thousand (10^-3)?
 
gibson101 said:
So to get to joules of work, you have to multiply pascals times volume? And why is the answer divided by a thousand (10^-3)?

Because the volume should be in cubic meters, and the values given in liters. How many liters in a cubic meter?
 
1 liter = .001 m^3. So pascals times cubic meters equals joules? I'm confused.
 
gibson101 said:
1 liter = .001 m^3. So pascals times cubic meters equals joules? I'm confused.

No need to be confused. Pressure x volume does indeed have the units of energy.
 

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