A system goes from A to B via two processes I and II as shown in figure- If -U1 and -U2 are the changes in internal energies in the processes I and II respectively- then-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Solubility Product

A system goes...

Question

A system goes from $A$ to $B$ via two processes $I$ and $I I$ as shown in figure. If $Δ U_{1}$ and $Δ U_{2}$ are the changes in internal energies in the processes $I$ and $I I$ respectively, then:

relation between

Δ U_{1}

and

Δ U_{2}

cannot be determined

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ U_{1} = Δ U_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Δ U_{1} < Δ U_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ U_{1} > Δ U_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $Δ U_{1} = Δ U_{2}$
Internal Energy (U) is a state function. It's value depends only on the initial and final positions.
Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e
$Δ U_{1} = Δ U_{2}$

Suggest Corrections

Similar questions

Q. A system goes from state 'A' to 'B' via two processes I and II as shown in figure. If

Δ U_{1}

and

Δ U_{2}

are the change in internal energies in the processes I and II respectively, then

Q. Figure shows two processes

a

and

b

for a given sample of a gas. If

Δ Q_{1}, Δ Q_{2}

are the amounts of heat absorbed by the system in the two cases and

Δ U_{1}, Δ U_{2}

are changes in internal energies respectively, then

Q. An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. If

Q_{1}, Q_{2}, Q_{3}

indicate the heat absorbed by the gas along the three processes and

Δ U_{1}, Δ U_{2}, Δ U_{3}

indicate the change in internal energy along the three processes respectively, then

A gas is expanded from volume $V_{0} t o 2 V_{0}$ under three different processes, as shown in the figure. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let $Δ U_{1}, Δ U_{2}$ and $Δ U_{3}$ be the change in the internal energy of the gas in these three processes, then:

Q. Let

Δ U_{1}

and

Δ U_{2}

be the change in internal energy in process A and B respectively,

Δ Q

be the net heat given to the system in process (A+B) and

Δ W

be the net work done by the system in the process (A+ B).
i)

Δ U_{1} + Δ U_{2} = 0

ii)

Δ U_{1} - Δ U_{2} = 0

iii)

Δ Q + Δ W = 0

iv)

Δ Q + Δ W = 0

Join BYJU'S Learning Program

A system goes from A to B via two processes I and II as shown in figure. If ΔU1 and ΔU2 are the changes in internal energies in the processes I and II respectively, then:

The correct option is B ΔU1=ΔU2Internal Energy (U) is a state function. It's value depends only on the initial and final positions.Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.eΔU1=ΔU2

A system goes from $A$ to $B$ via two processes $I$ and $I I$ as shown in figure. If $Δ U_{1}$ and $Δ U_{2}$ are the changes in internal energies in the processes $I$ and $I I$ respectively, then:

The correct option is B $Δ U_{1} = Δ U_{2}$
Internal Energy (U) is a state function. It's value depends only on the initial and final positions.
Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e
$Δ U_{1} = Δ U_{2}$