Question

A Carnot engine employs 1.5 mol of nitrogen gas as a working substance, which is considered as an ideal diatomic gas with γ=7/5 at the working temperatures of the engine. The Carnot cycle goes in the cycle ABCDA with AB being an isothermal expansion. The volume at points A and C of the cycle are 5.0×10−3m³ and 0.15 L, respectively. The engine operates between two thermal baths of temperature 500 K and 300 K.

(a) Find the values of volume at B and D.

b) How much heat is absorbed by the gas in the AB isothermal expansion?

c) How much work is done by the gas in the AB isothermal expansion?

d) How much heat is given up by the gas in the CD isothermal expansion?

Answer 1

(a)
`\(V_B = 2.0 * 10^(-3) m^3\) and \(V_D = 6.0 * 10^(-3) m^3\)`

(b)
`\(Q_(AB) = 450 J\)`

(c)
`\(W_(AB) = 225 J\)`

(d)
`\(Q_(CD) = -450 J\)`

Step-by-step explanation:

a: The Carnot engine undergoes a cycle ABCDA, where AB is an isothermal expansion. Using the ideal gas law and the fact that AB is isothermal, we can find
`\(V_B\) and \(V_D\)`.

The volumes at A and C are given, and the temperatures at the two ends are known. Thus, the gas law \(PV = nRT\) helps calculate the volumes.

b: For the heat absorbed during the isothermal expansion AB, the formula
`\(Q = nC_v\Delta T\)`is used, where \(C_v\) is the molar heat capacity at constant volume.

The temperature change \(\Delta T\) during AB can be found from the temperatures of the two thermal baths.

c: The work done during AB, \(W_{AB}\), is given by the equation \(W = nRT\ln(V_B/V_A)\), where
`\(V_A\)` is the initial volume. The negative sign indicates work done on the gas.

d: During the CD isothermal expansion, the gas releases heat
`(Q_(CD)\)),` which is equal in magnitude but opposite in sign to the heat absorbed in AB due to the second law of thermodynamics. The work done during CD is analogous to AB but with different initial and final volumes.