Question

one mole of an ideal gas initially kept in a cylinder at pressure 1MPa and temperature 27 c made to expand until it's volume is3
1 how much work is done if the expansion is adiabatic, isobaric, isothermal​

Answer 1

Answer:

Solution:- (C) 150K

From first law of thermodynamics,

ΔU=q+W

As we know that, for adiabatic condition,

q=0

∴ΔU=W.....(1)

At constant volume,

ΔU=nC

v

ΔT.....(2)

From eq

n

(1)&(2), we have

W=nC

v

ΔT.....(3)

Given:-

T

i

=27℃=(27+273)K=300K

T

f

=T(say)=?

ΔT=T

f

−T

i

=(T−300)

C

v

=20J/K−mol

W=−3kJ=−3×10

3

J[∵Work done by the gas is negative]

Substituting all these values in eq

n

(3), we have

−3000=1×20×(T−300)

T−300=−150

⇒T=300−150=150K

Hence the final temperature is 150K

Answer 2

Answer:

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Explanation:

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