when one mole of a monoatomic ideal gas at T kelvin undergoes adiabatic change under a constant external pressure of 1 atm changes volume from 1 to 2 litres. The final temperatue in kelvin is?
answer:-
T+ { 2 / (3*0.0821)}
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Perhaps there are more than one way of doing this.
Using the first law of Thermodynamics : dQ = dU + P dV
dQ = 0 as no heat is given to the system or taken out from the system.
dU = μ C_v dT and Hence,
μ C_v dT = - P dV
As pressure is constant, μ C_v ΔT = - P ( V2 - V1 )
C_v = 3/2 R for a monoatomic gas
R = 8.314 J/mol/K μ = 1 mole V2 = 2 * 10⁻³ m³ V1 = 1 * 10⁻³ m³
P = 1.013 * 10⁵ N/m²
ΔT = 8.12 ° K
this seems to be same as 2 / (3*0.0821)
Using the first law of Thermodynamics : dQ = dU + P dV
dQ = 0 as no heat is given to the system or taken out from the system.
dU = μ C_v dT and Hence,
μ C_v dT = - P dV
As pressure is constant, μ C_v ΔT = - P ( V2 - V1 )
C_v = 3/2 R for a monoatomic gas
R = 8.314 J/mol/K μ = 1 mole V2 = 2 * 10⁻³ m³ V1 = 1 * 10⁻³ m³
P = 1.013 * 10⁵ N/m²
ΔT = 8.12 ° K
this seems to be same as 2 / (3*0.0821)
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