Timeline for What would be the enthalpy change for a isothermal expansion?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2020 at 12:46 | comment | added | Chet Miller | Yes. That is correct. | |
| Jun 12, 2020 at 12:03 | comment | added | maverick | @ChetMiller And that useful work done by system is same as the isothermal work which is either reversible or irreversible?? | |
| Jun 12, 2020 at 11:55 | comment | added | maverick | @Chet Miller just one more thing, is it right that ,in isothermal process, if $Q$ heat is absorbed from surrounding , then all the heat is converted into useful work done by system (Gibb's energy) and it has no contribution in enthalpy change of system?? | |
| Jun 12, 2020 at 11:46 | comment | added | Chet Miller | @maverick It is caused by the heat that is absorbed (i.e., transferred to the gas from its surroundings). | |
| Jun 12, 2020 at 7:00 | comment | added | maverick | @ChetMiller I don't understand if $\Delta H$ is 0 then which energy causes the isothermal work ? | |
| Sep 29, 2018 at 1:23 | comment | added | Chet Miller | If the initial and final equilibrium temperatures are the same (irrespective of what happens along the path), the change in enthalpy of an ideal gas is zero. The enthalpy and internal energy of an ideal gas are functions only of temperature. | |
| Sep 28, 2018 at 6:39 | comment | added | SmarthBansal | @ChesterMiller If I have an ideal gas, then will the change in enthalpy be zero, for an isothermal path? Does it matter whether the process is carried out reversibly or not? | |
| Apr 22, 2017 at 15:10 | vote | accept | Nitro phenol | ||
| Aug 18, 2016 at 13:14 | comment | added | Chet Miller | @PrateekChauhan For a single phase ideal gas, of course it is. | |
| Aug 18, 2016 at 3:33 | comment | added | Nitro phenol | @chester Miller if it were so then all isothermal process would be isoenthalpic which is not the case | |
| Aug 17, 2016 at 17:02 | vote | accept | Nitro phenol | ||
| Mar 29, 2017 at 4:30 | |||||
| Aug 17, 2016 at 13:17 | comment | added | Chet Miller | Yes. The enthalpy of an ideal gas is a function only of temperature, no matter what kind of process is imposed. | |
| Aug 17, 2016 at 13:12 | history | answered | user7951 | CC BY-SA 3.0 |