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Consider two moles of an ideal diatomic gas at 300 K and 0.507 MPa to be expanded adiabatically to a final pressure of 0.2033 MPa against a constant pressure of 0.101 MPa the final temperature is cv=1.5R?
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Consider two moles of an ideal diatomic gas at 300 K and 0.507 MPa to ...
Adiabatic Expansion of Ideal Diatomic Gas

Given


  • Two moles of an ideal diatomic gas

  • Initial temperature (T1) = 300 K

  • Initial pressure (P1) = 0.507 MPa

  • Final pressure (P2) = 0.2033 MPa

  • Constant pressure (Pext) = 0.101 MPa

  • Final temperature (T2) = ?

  • Specific heat capacity at constant volume (cv) = 1.5R



Calculations

Since the process is adiabatic, no heat is exchanged between the system and the surroundings, so Q = 0.

For an ideal gas, the equation of state is given by:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Since the process is reversible, we can use:

P1V1 = P2V2

Also, since the process is adiabatic, we have:

P1V1c = P2V2c

where c is the specific heat capacity at constant volume.

Combining the two equations:

P1V1c = P2V2c

P1V1c = P2(P1/P2)c/(c-1)V1c

Simplifying:

V2 = V1(P1/P2)1/c

Substituting into the equation of state:

P1V1 = nRT1

P2V2 = nRT2

we get:

P1V1 = P2V2(T2/T1)
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Consider two moles of an ideal diatomic gas at 300 K and 0.507 MPa to ...
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Consider two moles of an ideal diatomic gas at 300 K and 0.507 MPa to be expanded adiabatically to a final pressure of 0.2033 MPa against a constant pressure of 0.101 MPa the final temperature is cv=1.5R?
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