Question

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

Answer 1

Answer:

Explanation:

The energy for an isothermal expansion can be computed as:

$\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}$ --- (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:

$\mathtt{e = 1 - \dfrac{T_L}{T_H}}$

$e = \dfrac{T_H-T_L}{T_H}$

In addition to that, for any heat engine, the efficiency e =$\dfrac{W}{Q_H}$

relating the above two equations together, we have:

$\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}$

Making the work done (W) the subject:

$W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)$

From equation (1):

$\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}$

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

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:

$\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)$

$T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}$

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:

$\Big(\dfrac{V_c}{V_b}\Big) = 5.7$

∴

$T_H = T_L (5.7)^{y-1}$

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:

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

$\mathsf{930 = 15.7966\times 1.5315 (T_L )})$

$\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}$

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

From $T_H = T_L (5.7)^{0.67}$

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

$\mathbf{T_H \simeq 125K}$