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Standard XII

Chemistry

Work Done in Isothermal Reversible Process

Compute the h...

Question

Compute the heat added, the work done, and the change in internal energy if this is done at constant volume.

Δ Q = 199800 J

;

Δ W = 0 J

;

Δ U = 199800 J

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Δ Q = 199900 J

;

Δ W = 0 J

;

Δ U = 199800 J

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Δ Q = 199800 J

;

Δ W = 0 J

;

Δ U = 199900 J

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

Δ Q = 199900 J

;

Δ W = 0 J

;

Δ U = 199900 J

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

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Solution

The correct option is B $Δ Q = 199800 J$ ; $Δ W = 0 J$ ; $Δ U = 199800 J$
Initial temperature of gas $T_{1} = 10^{o} C$
Final temperature $T_{2} = 100^{o} C$
Mass of gas $m = 3 k g$
Change in internal energy $Δ U = n C_{v} Δ T$
Since $C_{v}$ is given in terms of $J k g^{- 1} L^{- 1}$ ,
$∴$ $Δ U = 3 \times C_{v} Δ T = 3 \times 740 \times (100 - 10) = 199800 J$
Calculation of work done :
Since V is constant, thus no work is done i.e. $Δ W = 0$ .
Calculation of heat :
From 1st law of thermodynamics,
$Δ Q = Δ U + Δ W$
$Δ Q = 199800 + 0 = 199800 J$

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Similar questions

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. 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

Q. Refer to figure. Let ∆U₁ and ∆U₂ be the change in internal energy in processes A and B respectively, ∆Q be the net heat given to the system in process A + B and ∆W be the net work done by the system in the process A + B.

(a) ∆U₁ + ∆U₂ = 0.
(b) ∆U₁ − ∆U₂ = 0.
(c) ∆Q − ∆W = 0.
(d) ∆Q + ∆W = 0

Q. An ideal gas undergoes the process

1 ⟶ 2

as shown in the figure, the heat supplied and work done in process is

Δ Q

and

Δ W

respectively. the ratio

Δ Q : Δ W

Q. The figures given below show different processes (relating pressure

P

and volume

V

) for a given amount for an ideal gas.

Δ W

is work done by the gas and

Δ Q

is heat absorbed by the gas.

\begin{matrix} Column I & Column II 1. In Fig (i) & (p) Δ Q > 0 2. In Fig (i i) & (q) Δ W < 0 3. In Fig (i i i) & (r) Δ Q < 0 4. In Fig (i v) & (s) Δ W > 0 \end{matrix}

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