Question

3.10 mol of an ideal gas with CV,m=3R/2 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure are Ti=305K and Pi=14.7bar. The final pressure is Pf=1.00bar. Calculate q, w, ΔU, ΔH for the process.

Answer 1

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:

$(\frac{P_1}{P_2})^{1-\gamma}=(\frac{T_2}{T_1})^{\gamma}$

where,

$P_1$ = initial pressure = 14.7 bar

$P_2$ = final pressure = 1.00 bar

$T_1$ = initial temperature = 305 K

$T_2$ = final temperature = ?

$\gamma=\frac{R}{C_v}+1=frac{R}{(\frac{3R}{2})}+1=\frac{5}{3}=1.67$

Now put all the given values in the above expression, we get:

$(\frac{14.7}{1.00})^{1-1.67}=(\frac{T_2}{305})^{1.67}$

$T_2=103.7K$

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,

$dq=dU-w\\\\0=dU-w\\\\dU=w\\\\w=dU=n\times C_v\times (T_2-T_1)$

$w=dU=3.10mol\times \frac{3R}{2}\times (103.7-305)K$

$w=dU=3.10mol\times \frac{3\times 8.314J/mol.K}{2}\times (103.7-305)K$

$w=dU=-7782.278J=-7.78kJ$

Now we have to calculate the value of $C_p$

$C_p-C_v=R\\\\C_p=R+C_v\\\\C_p=R+\frac{3R}{2}\\\\C_p=\frac{5R}{2}$

Now we have to calculate the value of ΔH.

$dH=n\times C_p\times (T_2-T_1)$

$dH=3.10mol\times \frac{5R}{2}\times (103.7-305)K$

$dH=3.10mol\times \frac{5\times 8.314J/mol.K}{2}\times (103.7-305)K$

$dH=-12970.5J=-12.97kJ$

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.