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Standard XII

Physics

Work

Which of the ...

Question

Which of the following options is the only correct representation of a process in which $Δ U = Δ Q - P Δ V$ ?

(I I I) (i i i) (P)

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(I I) (i i i) (P)

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(I I) (i i i) (S)

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(I I) ((i v) (R)

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Solution

The correct option is B $(I I) (i i i) (P)$
I: Work done by adiabatic process:

$W_{1 - 2} = \frac{(P_{2} V_{2} - P_{1} V_{1}}{γ - 1})$

II: Work done by isobaric process:
$W_{1 - 2} = - P V_{2} + P V_{1} = P Δ V$

III : Work done by isochoric Process :

$W_{1 - 2} = 0$ . As the volume is constant throughout the process.

IV: Work done by isothermal :
$W_{1 - 2} = - n R T l n \frac{V_{2}}{V_{1}}$

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Similar questions

Q. Assertion :In isothermal process, whole of the heat supplied to the body is converted into internal energy. Reason: According to the first law of thermodynamics:

Δ Q = Δ U + p Δ V

Choose the correct option.

An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding $P - V$ diagrams in column 3 of the table. Consider only the path from state I to state $2 .$ $W$ denotes the corresponding work done on the system. The equation and plots in the table have standard notations as used in thermodynamic processes. Here $γ$ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is $n .$

Column 1	Column 2	Column 3
(I) $W_{1 \to 2} = \frac{1}{γ - 1} (P_{2} V_{2} - P_{1} V_{1})$	(i) Isothermal	(P)
(II) $W_{1 \to 2} = - {P V}_{2} + {P V}_{1}$	(ii) Isochoric	(Q)
(III) $W_{1 \to 2} = 0$	(iii) Isobaric	(R)
(IV) $W_{1 \to 2} = - n R T ln (\frac{V_{2}}{V_{1}})$	(iv) Adiabatic	(S)

Which of the following options is the only correct representation of a process in which

Δ U = Δ Q - P Δ V ?

Q. A student records

Δ Q, Δ U, Δ W

for a thermodynamic cycle

A B C A

. Certain entries are missing. Find correct entry in following options.

Q. Match the thermodynamic processes taking place in a system with the correct conditions.
In the table,

Δ Q

is the heat supplied,

Δ W

is the work done and

Δ U

is the change in internal energy of the system.

Process	Condition
(i) Adiabatic	(A) $Δ W = 0$
(ii) Isothermal	(B) $Δ Q = 0$
(iii) Isochoric	(C) $Δ U \neq 0$ , $Δ W \neq 0$ , $Δ Q \neq 0$
(iv) Isobaric	(D) $Δ U = 0$

Q. Column - 1: Trigonometric expression
Column - 2: Dependency on '

α

' and '

β

'
Column - 3: Equivalent value of Trigonometric expression in Column - 1.

\begin{matrix} C o l u m n - 1 & C o l u m n - 2 & C o l u m n - 3 (I) & {cos}^{2} (α + β) - {sin}^{2} (α - β) - cos (2 α) cos (2 β) + 1 & (i) & Independent of α only & (P) & 1 (I I) & {cos}^{2} α + {cos}^{2} (α + β) - 2 cos (α) cos (β) cos (α + β) & (i i) & Independent of β only & (Q) & {sin}^{2} β (I I I) & \frac{sin (α - β) sin (α + β)}{1 - {tan}^{2} α {cot}^{2} β} & (i i i) & Independent of α and β & (R) & - {cos}^{2} α {sin}^{2} β (I V) & 2 {sin}^{2} β + 4 cos (α + β) sin α sin β + cos 2 (α + β) & (S) & cos (2 α) \end{matrix}

Which of the following options is the only correct combination ?

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Which of the following options is the only correct representation of a process in which ΔU=ΔQ−PΔV?

Which of the following options is the only correct representation of a process in which $Δ U = Δ Q - P Δ V$ ?