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

Byju's Answer

Standard XII

Physics

Specific Heat Capacity

Work done in ...

Question

Work done in reversible adiabatic process by an ideal gas is given by:

2.303 R T l o g

\frac{V_{1}}{V_{2}}

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

\frac{n R}{γ - 1} (T_{2} - T_{1})

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

2.303 R T l o g

\frac{V_{2}}{V_{1}}

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

none of these

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

Open in App

Solution

The correct option is A $\frac{n R}{γ - 1} (T_{2} - T_{1})$
Solution:- (D) $\frac{n R}{γ - 1} (T_{2} - T_{1})$
From first law of thermodynamics,
$Δ U = q + W$
As e know that, for an adiabatic process,
$q = 0$
$∴ Δ U = W . . . . . (1)$
As we know that,
$Δ U = n C_{v} Δ T$
where as,
$Δ T =$ Change in temperature $= T_{2} - T_{1}$
$C_{v} =$ heat capacity at constant volume $= \frac{R}{γ - 1}$
$n =$ no. of moles
$∴ Δ U = n \frac{R}{γ - 1} (T_{2} - T_{1}) . . . . . (2)$
From ${e q}^{n} (1) & (2)$ , we have
$W = \frac{n R}{γ - 1} (T_{2} - T_{1})$

Suggest Corrections

Similar questions

Q. A gas with

\frac{C_{P}}{C_{V}} = γ

goes from an initial state

(p_{1}, V_{1}, T_{1})

to a final state

(p_{2}, V_{2}, T_{2})

through an adiabatic process. The work done by the gas is

For an ideal gas, an illustration of three different paths

A

(B + C)

and

(D + E)

from an initial state

P_{1}

V_{1}

T_{1}

to a final state

P_{2}

V_{2}

T_{1}

is shown in the given figure.
Path

A

represents a reversible Isothermal expansion from

P_{1}

V_{1}

P_{2}, V_{2}

. Path

(B + C)

reperesents a reversible adiabatic expansion

(B)

from

P_{1} V_{1}

T_{1}

P_{3}

V_{2}

T_{2}

followed by reversible heating the gas at constant volume

(C)

from

P_{3}

V_{2}

T_{2}

P_{2}

V_{2}

T_{1}

. Path

(D + E)

represents a reversible expansion at constant pressure

P_{1} (D)

from

P_{1}

V_{1}

T_{1}

P_{1}

V_{2}

T_{3}

followed by a reversible cooling at constant volume

V_{2} (E)

from

P_{1}

V_{2}

T_{3}

P_{2}

V_{2}

T_{1}

. What is

Δ S

for path

(D + E) ?

Q. An ideal gas is expanded from

p_{1}, V_{1}, T_{1}

p_{2}, V_{2}, T_{2}

under different conditions.the correct statement(s) among the following is(are)

Q. For an ideal gas, an illustration of three different paths

A

(B + C)

and

(D + E)

from an initial state

P_{1}

V_{1}

T_{1}

to a final state

P_{2}

V_{2}

T_{1}

is shown in the given figure.
Path A represents a reversible Isothermal expansion from

P_{1}

V_{1} t o P_{2}, V_{2}

. Path

(B + C)

reperesents a reversible adiabatic expansion

(B)

from

P_{1} V_{1}

T_{1} t o P_{3}

V_{2}

T_{2}

followed by reversible heating the gas at constant volume (C) from

P_{3}

V_{2}

T_{2} t o P_{2}

V_{2}

T_{1}

. Path ?

(D + E)

represents a reversible expansion at constant pressure

P_{1} (D)

from

P_{1}

V_{1}

T_{1}

P_{1}

V_{2}

T_{3}

followed by a reversible cooling at constant volume

V_{2} (E)

from

P_{1}

V_{2}

T_{3} t o P_{2}

V_{2}

T_{1}

What is

q_{r e v}

, for path

(A)

Q. Starting with the same initial conditions, one mole of an ideal monoatomic gas expands reversibly from volume

V_{1}

V_{2}

in two different ways. The work done by gas is

w_{1}

, if the process is purely isothermal,

w_{2}

if purely isobaric and

w_{3}

if purely adiabatic, then:

Join BYJU'S Learning Program