Question

One kg of air contained in a piston–cylinder assembly undergoes a process from an initial state where T1 5 300 K, 1 5 0.8 m3/kg to a final state where T2 5 420 K, 2 5 0.2 m3/kg. Can this process occur adiabatically? If yes, determine the work, in kJ, for an adiabatic process between these states. If no, determine the direction of the heat transfer. Assume the ideal gas model for air.

Answer 1

Answer:

Given that

T ₁= 300 K, V₁ = 0.8 ³m/kg x 1kg = 0.8 m³

         T₂ = 420 K, V₂ = 0.2 m³/kg x 1kg =0.2 m³

For adiabatic

$PV^{\gamma} = C\\TV^{\gamma-1} = C$

For air γ = 1.4 so, γ-1 = 0.4

$T_1V_1^{\gamma-1} =300\times 0.80.4 = 274.3830$

$T_2V_2^{\gamma-1} =420\times 0.20.4 = 220.6283$

This is not equal so the process is not adiabatic.

We know that

PV = mRT

P₁ x 0.8 = 1x 0.287  x 300

P₁ = 107.625 kPa

P₂x 0.2= 1 x 0.287 x 420

P₂= 602.7 kPa

The pressure of the system is increase it means that work is done on the system so the heat will transfer form the system.

Answer 2

Answer:

This process is not adiabatic.

Direction: System to surrounding.

Explanation:

The change of entropy is:

$delta S=Cvln\frac{T2}{T1}+Rln\frac{V2}{V1}$

Replacing:

$deltaS=0.718ln\frac{420}{300}+0.287ln\frac{0.2}{0.8}=-0.16 kJ/kg$

As seen in the result, entropy is negative, therefore, the system is losing heat. Therefore, this process is not adiabatic.

The direction of heat transfer is:

$Q=Cp(T2-T1)=1.005(420-300)=120.6 kJ/kg$

Direction: System to surrounding.