Answer The value of
and
is, 40.79 kJ and 37.7 kJ respectively.
Explanation :
Heat released at constant pressure is known as enthalpy.
The formula used for change in enthalpy of the gas is:
![\Delta Q_p=\Delta H=40.79kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20Q_p%3D%5CDelta%20H%3D40.79kJ%2Fmol)
Now we have to calculate the work done.
Formula used :
![w=-P\Delta V\\\\w=-P\times (V_2-V_1)](https://tex.z-dn.net/?f=w%3D-P%5CDelta%20V%5C%5C%5C%5Cw%3D-P%5Ctimes%20%28V_2-V_1%29)
where,
w = work done = ?
P = external pressure of the gas = 1 atm
= initial volume = ![1.88\times 10^{-5}m^3=1.88\times 10^{-5}\times 10^3L=1.88\times 10^{-2}L](https://tex.z-dn.net/?f=1.88%5Ctimes%2010%5E%7B-5%7Dm%5E3%3D1.88%5Ctimes%2010%5E%7B-5%7D%5Ctimes%2010%5E3L%3D1.88%5Ctimes%2010%5E%7B-2%7DL)
= final volume = ![3.06\times 10^{-2}m^3=3.06\times 10^{-2}\times 10^3L=3.06\times 10^{1}L](https://tex.z-dn.net/?f=3.06%5Ctimes%2010%5E%7B-2%7Dm%5E3%3D3.06%5Ctimes%2010%5E%7B-2%7D%5Ctimes%2010%5E3L%3D3.06%5Ctimes%2010%5E%7B1%7DL)
Now put all the given values in the above formula, we get:
![w=-(1atm)\times (3.06\times 10^{1}-1.88\times 10^{-2})L](https://tex.z-dn.net/?f=w%3D-%281atm%29%5Ctimes%20%283.06%5Ctimes%2010%5E%7B1%7D-1.88%5Ctimes%2010%5E%7B-2%7D%29L)
![w=-30.5812L.atm=-30.5812\times 101.3J=-3097.87556J=-3.09\times 10^3J=-3.09kJ](https://tex.z-dn.net/?f=w%3D-30.5812L.atm%3D-30.5812%5Ctimes%20101.3J%3D-3097.87556J%3D-3.09%5Ctimes%2010%5E3J%3D-3.09kJ)
Now we have to calculate the change in internal energy.
![\Delta U=q+w](https://tex.z-dn.net/?f=%5CDelta%20U%3Dq%2Bw)
![\Delta U=40.79kJ+(3.09kJ)](https://tex.z-dn.net/?f=%5CDelta%20U%3D40.79kJ%2B%283.09kJ%29)
![\Delta U=37.7kJ](https://tex.z-dn.net/?f=%5CDelta%20U%3D37.7kJ)
Thus, the value of
and
is, 40.79 kJ and 37.7 kJ respectively.