Answer:
The change in temperature is -141.428K
The entropy change is -60.75J/mole K Kg
Explanation:
Given that,
Initial Temperature, T₁ = 330k
Initial Pressure, P₁ = 3500kpa
Final pressure, P₂ = 2000kpa
Since the work done on the surrounding is zero
hence, the change in volume of the system will be zero and the process will be constant
Calculate the Temperature of the system
[tex]\frac{P_1}{P_2} = \frac{T_1}{T_2}[/tex]
Substitute the value in the above equation
[tex]\frac{3500}{2000} = \frac{330}{T_2} \\\\T_2 = 188.57K[/tex]
Calculate the change in temperature
ΔT = T₂ - T₁
= 188.57K - 330K
= -141.428K
The change in temperature is -141.428K
Expression of change in entropy
[tex]ds =\frac{dQ}{T}[/tex]
ds is the entropy change
equation for [tex]dQ = c_vdT[/tex]
Substitute for dQ in the expression of change in entropy and integrate
[tex]ds = \frac{C_vdT}{T}[/tex]
[tex]\int\limits^ {s_2 }_ {s_1 } \, ds = \int\limits^ { T_2}_ {T_1 } \frac{C_vdT}{T}[/tex]
[tex]s_2-s_1=C_vIn\frac{T_2}{T_1}[/tex]
substitute the values
[tex]\Delta s = \frac{152}{1.4} In\frac{188.57}{330} \\\\\Delta s = -60.75J/Mole-K -Kg[/tex]
The entropy change is -60.75J/mole K Kg
