An ideal diatomic gas has an initial pressure of 1.00×10°Pa , an initial volume of 2.00m° , and an initial temperature of 300K. (This is point 1 on the pV-diagram.) The gas has an isochoric increase in pressure to 2.00×10° Pa. (This is point 2 on the pV-diagram.) The gas then has an isothermal expansion to a volume of 3.00m' . (This is point 3 on the pV-diagram.) The pressure is then reduced adiabatically back down to its original pressure of 1.00×10°P.. (This is point 4 on the pV-diagram.) Finally, the gas has an isobaric decrease in volume to its original volume of 2.00m. (The gas is back to point 1 on the pV-diagram.) Fill in the missing values on the following table. а. Point Volume Pressure. Temnerature
An ideal diatomic gas has an initial pressure of 1.00×10°Pa , an initial volume of 2.00m° , and an initial temperature of 300K. (This is point 1 on the pV-diagram.) The gas has an isochoric increase in pressure to 2.00×10° Pa. (This is point 2 on the pV-diagram.) The gas then has an isothermal expansion to a volume of 3.00m' . (This is point 3 on the pV-diagram.) The pressure is then reduced adiabatically back down to its original pressure of 1.00×10°P.. (This is point 4 on the pV-diagram.) Finally, the gas has an isobaric decrease in volume to its original volume of 2.00m. (The gas is back to point 1 on the pV-diagram.) Fill in the missing values on the following table. а. Point Volume Pressure. Temnerature
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part b please
![An ideal diatomic gas has an initial pressure of 1.00×10°Pa , an initial volume of 2.00m', and
an initial temperature of 300K. (This is point 1 on the pV-diagram.) The gas has an isochoric
increase in pressure to 2.00×10°P. . (This is point 2 on the pV-diagram.) The gas then has an
isothermal expansion to a volume of 3.00m'. (This is point 3 on the pV-diagram.) The pressure
is then reduced adiabatically back down to its original pressure of 1.00×10°P.. (This is point 4
on the pV-diagram.) Finally, the gas has an isobaric decrease in volume to its original volume of
2.00m. (The gas is back to point 1 on the pV-diagram.)
а.
Fill in the missing values on the following table.
Point
Volume,
Pressure,
Temperature,
v (m²)
p(10ʻPA)
T(K)
1
2.00
1.00
300
2
2.00
3
3.00
4
1.00
b. Fill in the values for each of the processes in the following table. (These values
correspond to the First Law of Thermodynamics written as: AE, =W +Q.)
Process
Change in internal
Work done to gas,
Heat added to gas,
energy, AE, (J)
W (3)
Q(1)
'th
1→ 2
2 → 3
3 → 4
4 →1
Full Cycle](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdedc6ff1-e5e7-48e1-8e69-1d621b5c4e4e%2Fec1a1cb9-88d4-48ce-a0ac-507552699316%2Fme2komzh_processed.png&w=3840&q=75)
Transcribed Image Text:An ideal diatomic gas has an initial pressure of 1.00×10°Pa , an initial volume of 2.00m', and
an initial temperature of 300K. (This is point 1 on the pV-diagram.) The gas has an isochoric
increase in pressure to 2.00×10°P. . (This is point 2 on the pV-diagram.) The gas then has an
isothermal expansion to a volume of 3.00m'. (This is point 3 on the pV-diagram.) The pressure
is then reduced adiabatically back down to its original pressure of 1.00×10°P.. (This is point 4
on the pV-diagram.) Finally, the gas has an isobaric decrease in volume to its original volume of
2.00m. (The gas is back to point 1 on the pV-diagram.)
а.
Fill in the missing values on the following table.
Point
Volume,
Pressure,
Temperature,
v (m²)
p(10ʻPA)
T(K)
1
2.00
1.00
300
2
2.00
3
3.00
4
1.00
b. Fill in the values for each of the processes in the following table. (These values
correspond to the First Law of Thermodynamics written as: AE, =W +Q.)
Process
Change in internal
Work done to gas,
Heat added to gas,
energy, AE, (J)
W (3)
Q(1)
'th
1→ 2
2 → 3
3 → 4
4 →1
Full Cycle
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