Respuesta :
Answer : The value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.
Explanation :
First we have to calculate the final temperature of the gas.
Using poission equation:
[tex](\frac{P_1}{P_2})^{1-\gamma}=(\frac{T_2}{T_1})^{\gamma}[/tex]
where,
[tex]P_1[/tex] = initial pressure = 14.7 bar
[tex]P_2[/tex] = final pressure = 1.00 bar
[tex]T_1[/tex] = initial temperature = 305 K
[tex]T_2[/tex] = final temperature = ?
[tex]\gamma=\frac{R}{C_v}+1=frac{R}{(\frac{3R}{2})}+1=\frac{5}{3}=1.67[/tex]
Now put all the given values in the above expression, we get:
[tex](\frac{14.7}{1.00})^{1-1.67}=(\frac{T_2}{305})^{1.67}[/tex]
[tex]T_2=103.7K[/tex]
Now we have to calculate the q, w, ΔU and ΔH for the process.
As we know that, at adiabatic process the value of q=0. So,
[tex]dq=dU-w\\\\0=dU-w\\\\dU=w\\\\w=dU=n\times C_v\times (T_2-T_1)[/tex]
[tex]w=dU=3.10mol\times \frac{3R}{2}\times (103.7-305)K[/tex]
[tex]w=dU=3.10mol\times \frac{3\times 8.314J/mol.K}{2}\times (103.7-305)K[/tex]
[tex]w=dU=-7782.278J=-7.78kJ[/tex]
Now we have to calculate the value of [tex]C_p[/tex]
[tex]C_p-C_v=R\\\\C_p=R+C_v\\\\C_p=R+\frac{3R}{2}\\\\C_p=\frac{5R}{2}[/tex]
Now we have to calculate the value of ΔH.
[tex]dH=n\times C_p\times (T_2-T_1)[/tex]
[tex]dH=3.10mol\times \frac{5R}{2}\times (103.7-305)K[/tex]
[tex]dH=3.10mol\times \frac{5\times 8.314J/mol.K}{2}\times (103.7-305)K[/tex]
[tex]dH=-12970.5J=-12.97kJ[/tex]
Thus, the value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.
