Question

V In a thermodynamic process two moles of a
monatomic ideal gas obeys P « V-2. If temperature
of the gas increases from 300 K to 400 K, then find
work done by the gas (where R = universal gas
constant)​

Answer 1

Answer:

Given,

P∝V

−2

n=2

T

1

=300K

T

2

=400K

PV

2

=const. . . . . .(1)

γ=2 (PV

γ

=constant)

The above equation shows that the process is adiabatic process,

The work done in adiabatic process in thermodynamics is given by

W=nR

γ−t

(T

1

−T

2

)

W=nR

2−1

(300−400)

W=−100nR=−2×100R

W=−200R

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Answer 2

Explanation:

Correct option is B)

Given,

P∝V−2

n=2

T1=300K

T2=400K

PV2=const. . . . . .(1)

γ=2            (PVγ=constant)

The above equation shows that the process is adiabatic process,

The work done in adiabatic   process in thermodynamics is given by

W=nRγ−t(T1−T2)

W=nR2−1(300−400)

W=−100nR=−2×100R

W=−200R

The correct option is B.