Respuesta :
Answer:
[tex]T_{2} = 25^{\circ}C[/tex]
Solution:
As per the question:
Temperature, [tex]T_{1} = 25^{\circ}C[/tex]
Pressure, [tex]P_{1} = 5 atm[\tex]
Now,
We know that the process of throttling occurs in adiabatic conditions where no heat is transferred in between the system and the surrounding, i.e., Q = 0
Thus no work is done in this process, i.e., W = 0
Enthalpy also remains same from one state to the other state, i.e., [tex]\Delta h = 0[/tex]
Therefore, from the eqn:
[tex]Q - W = \Delta h + \Delta KE[/tex]
We can write:
[tex]\Delta h = h_{1} - h_{2} = 0[/tex]
[tex]\Delta h = C_{p}\Delta T[/tex]
[tex]C_{p}(T_{2} - T_{1}) = 0[/tex]
Thus
[tex]T_{1} = T_{2} = 25^{\circ}C[/tex]
where
[tex]T_{2}[/tex] = Exit temperature
