A sample of argon gas at 1 atm and 27°c expands reversibly and adiabatically from 1.25 ltre to 2.50 litre. the change in enthalpy approximately is (a) 11.5 joul

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can't we use dH = dU +pdV and find enthalpy?

im doing like this but not getting answers.. plz check my mistake

Adiabatic process me Pressure volume & temperature koi constant nahi hota

Got this sourav??

can't we use dH = dU +pdV

according to given data you use the formula na ...

yes sir, i m just asking if we can't use that formula

bcoz we also have all the data needed to solve that formula also

yes sourav

use formula according to all the data provided

got this??

im done like this.. im not getting answers.. plz check what is my mistake

sourav either directly Cv value is given to you or argon is monoatomic so Cv=3/2R you calculate it comes 12.48..now Cv is calculated...as relation Cp-Cv=R...you have to calculate Cp..as because heat capacity at constant volume relates with enthalpy...then delH= n Cp delT..you get the value

yea sir that i know.. but i m just curious why answer is not coming like the way i did

dear sourav If temperature and pressure remain constant through the process and the work is limited to pressure-volume work, then the enthalpy change is given by the equation del H= delU + P delV

here only at constant 1 atm pressure it occurs...

So directly relation of enthalpy with heat capacity at constant pressure is applicable

but sir if temp remains constant throughout the process the internal energy should be 0 na. then dH should be only pdV.. here pressure is constant only

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy read this article sourav

we are talking about standard temp and pressure conditions