Which is not the correct relation between ent | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The fundamental relation between enthalpy $(\Delta H)$ and internal energy $(\Delta E)$ for a process at constant pressure is given by $\Delta H =

Vedclass · Accepted Answer

$\Delta H = \Delta E - P \Delta V$

Vedclass · Answer

(C) The fundamental relation between enthalpy $(\Delta H)$ and internal energy $(\Delta E)$ for a process at constant pressure is given by $\Delta H = \Delta E + P \Delta V$. Rearranging this equation gives $\Delta E = \Delta H - P \Delta V$. For gaseous reactions,the work done $P \Delta V$ is equivalent to $\Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous species. Thus,$\Delta H = \Delta E + \Delta n_g RT$ (often written as $\Delta H = \Delta E + n RT$). Therefore,the relation $\Delta H = \Delta E - P \Delta V$ is incorrect.

Which is not the correct relation between enthalpy $(\Delta H)$ and internal energy $(\Delta E)$?

Similar Questions

The evaporation of water is a:

At $27\,^{\circ}C$,$1\, mole$ of an ideal gas expands reversibly and isothermally from $2\, atm$ to $1\, atm$. Then $\Delta H$ will be......$kJ$.

$2 \ \text{moles}$ of a gas undergoes a change in state from $(2 \ \text{bar}, 40 \ \text{L})$ to $(5 \ \text{bar}, 15 \ \text{L})$ at a constant temperature. If the change in internal energy is given by $35 \ \text{bar-L}$, then the enthalpy change for the process will be: (in $\text{bar-L}$)

Which of the following is correct regarding the internal energy of a substance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module