The total energy in a system due to the temperature and pressure per unit mass in that system is known as specific enthalpy. It is used in thermodynamic equations when one desires to know the energy for a given single unit mass of a component.
Specific enthalpy is calculated by the equation:
H = U + PV
in the given case, Specific volume = 4.684 cm³/g = 149.888 cm³/g moles = 149.888 × 10⁻³ J/g moles
Specific internal energy (U) is 1706 J/mol and pressure is 41.64.
H = 1706 + 41.64 × 149.888 × 10⁻³ × 101.3 joules
= 2428 joules / mole
Answer:
+279.744 kJ.
Below is an attachment containing the solution to question.
The pressure needed to change the volume to 24 L is 0.796 atm.
Explanation:
It is known by Boyle's law that the pressure experienced by the gas molecules will be inversely proportional to the volume occupied by the molecules.

So as the initial volume is said to be 22.4 L, consider it as V₁ = 22.4 L. Then the initial pressure is said to be 0.853 atm, so P₁ = 0.853 atm. So we have to determine the new pressure P₂ when the volume is changed to V₂ = 24 L. As there is increase in the volume, the pressure should be decreased due to Boyle's law. Thus, as per Boyle's law, the two pressures and their volumes can be related as


Thus, the pressure gets decreased to 0.796 atm on increase in the volume to 24 L.
So the pressure needed to change the volume to 24 L is 0.796 atm.
Answer:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)

