Question

urgent

if 1.0 Kcal of heat is added to 1.2 L of O2 in a cylinder of constant pressure of 1 atm ,the volume increases to 1.5 L. calculate delta U delta H of the process.

Answer 1

Answer : $\Delta U=992.737cal$ and $\Delta H=1000cal$

Solution : Given,

Heat = 1 Kcal = 1000 cal       (1 Kcal = 1000 cal)

Initial volume = 1.2 L

Final volume = 1.5 L

Pressure = 1 atm

Formula used :

$\Delta U=q-w\\\Delta U=q-P\Delta V\\\Delta U=q-P(V_{final}-V_{initial})$

Now put all the given values in this formula, we get

$\Delta U=1000cal-[1atm\times (1.5L-1.2L)]\times 24.21=992.737cal$   $(1Latm=24.21cal)$

Now we have to calculate the $\Delta H$.

At constant pressure, $\Delta H=q_p$

So, $\Delta H=q_p=1000cal$

Therefore, $\Delta U=992.737cal$ and $\Delta H=1000cal$

Answer 2

Answer : \Delta U=992.737calΔU=992.737cal and \Delta H=1000calΔH=1000cal

Solution : Given,

Heat = 1 Kcal = 1000 cal       (1 Kcal = 1000 cal)

Initial volume = 1.2 L

Final volume = 1.5 L

Pressure = 1 atm

Formula used :

\begin{lgathered}\Delta U=q-w\\\Delta U=q-P\Delta V\\\Delta U=q-P(V_{final}-V_{initial})\end{lgathered}ΔU=q−wΔU=q−PΔVΔU=q−P(Vfinal​−Vinitial​)​

Now put all the given values in this formula, we get

\Delta U=1000cal-[1atm\times (1.5L-1.2L)]\times 24.21=992.737calΔU=1000cal−[1atm×(1.5L−1.2L)]×24.21=992.737cal(1Latm=24.21cal)(1Latm=24.21cal)

Now we have to calculate the \Delta HΔH .

At constant pressure, \Delta H=q_pΔH=qp​

So, \Delta H=q_p=1000calΔH=qp​=1000cal

Therefore, \Delta U=992.737calΔU=992.737cal and \Delta H=1000calΔH=1000cal

your answer is ready........