A flexible box contains 5.60 grams of nitrogen gas (N2) which is maintained at a constant pressure of 1.35 x 105 Pa. The box is placed over a fire, causing the volume to increase from 0.00200 m³ to 0.00300 m³. Find the increase in temperature of the gas. (For N2 molar mass M = 28 grams.)
A flexible box contains 5.60 grams of nitrogen gas (N2) which is maintained at a constant pressure of 1.35 x 105 Pa. The box is placed over a fire, causing the volume to increase from 0.00200 m³ to 0.00300 m³. Find the increase in temperature of the gas. (For N2 molar mass M = 28 grams.)
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![A flexible box contains 5.60 grams of nitrogen gas (N2) which is maintained at a
constant pressure of 1.35 x 10$ Pa. The box is placed over a fire, causing the
volume to increase from 0.00200 m3 to 0.00300 m³. Find the increase in
temperature of the gas.
(For N2 molar mass M = 28 grams.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff456c65b-f482-4083-8c37-6fe7d8344e1a%2Ffc1f0b12-c300-4cbf-a5f9-f121eb2303ce%2Fwqfkrje_processed.png&w=3840&q=75)
Transcribed Image Text:A flexible box contains 5.60 grams of nitrogen gas (N2) which is maintained at a
constant pressure of 1.35 x 10$ Pa. The box is placed over a fire, causing the
volume to increase from 0.00200 m3 to 0.00300 m³. Find the increase in
temperature of the gas.
(For N2 molar mass M = 28 grams.)
![Formulas:
Work done on a gas: Wgas
= - PAV
First Law of Thermodynamics: AU = Q+ W
Idea gas law: PV = nRT, thus, PAV = NRAT
Molar specific heat at constant volume: Cv : AU = nCvAt where n is moles
3
R for monatomic gasses,
2
5
R for diatomic gasses
2
The value of Cv is just:
5
R for monatomic gasses,
2
7
for diatomic gasses
2
The value of Cp is just:
Joules
Where R is the universal gas constant: R = 8.31
mole-Kelvin
Molar specific heat at constant pressure: Q = CPAT
CP
For adiabatic processes:
PVY = constant, where
Y
Cv
Work done at constant temperature:
W = -nRT In(
1
The First Law and Thermodynamic Processes (Ideal Gases)
Process
AU
Q
W
Isobaric
nC, AT
nC, AT
-P AV
Adiabatic
nC, AT
Δυ
Isovolumetric
nC, AT
AU
Isothermal
-W
-nRT In](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff456c65b-f482-4083-8c37-6fe7d8344e1a%2Ffc1f0b12-c300-4cbf-a5f9-f121eb2303ce%2F3lavx1j_processed.png&w=3840&q=75)
Transcribed Image Text:Formulas:
Work done on a gas: Wgas
= - PAV
First Law of Thermodynamics: AU = Q+ W
Idea gas law: PV = nRT, thus, PAV = NRAT
Molar specific heat at constant volume: Cv : AU = nCvAt where n is moles
3
R for monatomic gasses,
2
5
R for diatomic gasses
2
The value of Cv is just:
5
R for monatomic gasses,
2
7
for diatomic gasses
2
The value of Cp is just:
Joules
Where R is the universal gas constant: R = 8.31
mole-Kelvin
Molar specific heat at constant pressure: Q = CPAT
CP
For adiabatic processes:
PVY = constant, where
Y
Cv
Work done at constant temperature:
W = -nRT In(
1
The First Law and Thermodynamic Processes (Ideal Gases)
Process
AU
Q
W
Isobaric
nC, AT
nC, AT
-P AV
Adiabatic
nC, AT
Δυ
Isovolumetric
nC, AT
AU
Isothermal
-W
-nRT In
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