- #1
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
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