Respuesta :
Answer:
The hot temperature is 157.5 K
The cold temperature is 48.8 K
Explanation:
Step 1: Data given
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.
The volume increases by a factor of 5.7
The work output of the engine is 940 J in each cycle.
During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)
Step 2:For a carnot engine:
V2/V1 = V4/V3
Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))
⇒with Work = the work done in the cycle = 940J
⇒with n = the number of moles = 1.50 moles
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = the hot temperature
⇒With T2⇒ the cold temperature
where R = 8.31 J/mol K Gas Constant
940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))
940 = 1.5 * 8.314 ln(2) * (T1-T2)
(T1-T2) = 940 / (1.5*8.314*ln(2))
(T1-T2) = 108.7K
For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)
For a monatomic ideal gas, Cv = 3/2R
When we combine both, we'll have:
T2 = T1*(1/5.7)^(R/3/2R)
T2 = T1*(1/5.7)^(2/3)
T2= T1 * 0.31
Since we know that (T1-T2) = 108.7K
we have:
T1 - 0.31T1= 108.7K
0.69T1 = 108.7K
T1 = 157.5K
T2 = 157.5*0.31 = 48.8K
