Why multiply atm by atmospheric pressure in pascals?

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Homework Help Overview

The discussion revolves around the calculation of work done by an ideal gas during a two-step process involving pressure changes and volume expansion. The context includes the conversion of pressure from atmospheres to pascals and the relationship between pressure, volume, and work in thermodynamics.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the necessity of converting pressure units for calculations, questioning why 1.4 atm is multiplied by 1.013e5 pascals. There are inquiries about the relationship between pressure, volume, and work, as well as the significance of unit conversions.

Discussion Status

Some participants have provided clarifications regarding unit conversions and the relationship between pressure and volume in the context of work. There is ongoing exploration of the implications of using different units and how they affect the calculation of work done by the gas.

Contextual Notes

Participants are discussing the conversion of liters to cubic meters and the implications of using these units in calculations. The original poster and others express confusion about the conversion factors and the resulting units of energy.

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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