Question

The working substance of a certain Carnot engine is 1.90 of an ideal

monatomic gas. During the isothermal expansion portion of this engine's
cycle, the volume of the gas doubles, while during the adiabatic expansion
the volume increases by a factor of 5.7. The work output of the engine is
930 in each cycle.
Compute the temperatures of the two reservoirs between which this engine
operates.

Answer 1

Step-by-step explanation:

The energy for an isothermal expansion can be computed as:

`\mathsf{Q_H =nRTIn ((V_b)/(V_a))}` --- (1)

However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:

`V_b = 2V_a`

Equation (1) can be written as:

`\mathtt{Q_H = nRT_H In (2)}`

Also, in a Carnot engine, the efficiency can be computed as:

`\mathtt{e = 1 - (T_L)/(T_H)}`

`e = (T_H-T_L)/(T_H)`

In addition to that, for any heat engine, the efficiency e =
`(W)/(Q_H)`

relating the above two equations together, we have:

`(T_H-T_L)/(T_H) = (W)/(Q_H)`

Making the work done (W) the subject:

`W = Q_H \Big((T_H-T_L)/(T_H) \Big)`

From equation (1):

`\mathsf{W = nRT_HIn(2) \Big((T_H-T_L)/(T_H) \Big)}`

`\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}`

If we consider the adiabatic expansion as well:

`PV^y` = constant

i.e.

`P_bV_b^y = P_cV_c^y`

From ideal gas PV = nRT

we can have:

`(nRT_H)/(V_b)(V_b^y)= (nRT_L)/(V_c)(V_c^y)`

`T_H = T_L \Big((V_c)/(V_b)\Big)^(y-1)`

From the question, let us recall aw we are being informed that:

If the volumes changes by a factor = 5.7

Then, it implies that:

`\Big((V_c)/(V_b)\Big) = 5.7`

∴

`T_H = T_L (5.7)^(y-1)`

In an ideal monoatomic gas
`\gamma = 1.6`

As such:

`T_H = T_L (5.7)^(1.6-1)`

`T_H = T_L (5.7)^(0.67)`

Replacing the value of
`T_H = T_L (5.7)^(0.67)` into equation
`\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}`

`\mathsf{W = nRT_L In(2) (5.7 ^(0.67 )-1}})`

From in the question:

W = 930 J and the moles = 1.90

using 8.314 as constant

Then:

`\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^(0.67 )-1}})`

`\mathsf{930 = 15.7966* 1.5315 (T_L )})`

`\mathsf{T_L= (930 )/(15.7966* 1.5315)}`

`\mathbf{T_L \simeq = 39 \ K}`

From
`T_H = T_L (5.7)^(0.67)`

`\mathsf{T_H = 39 (5.7)^(0.67)}`

`\mathbf{T_H \simeq 125K}`