Question

A gas is allowed to expand in a well insulated
container against a constant external pressure of
2.5 atm from an initial volume of 2.50 L to a final
volume of 4.50 L. The change in internal energy AU
of the gas in joules will be [NEET-2017]
(1) 1136.25 J (2) -500 J
(3) -505 J
(4) +505 J

Answer 1

Answer:

(3) -505 J

Explanation:

Given

pressure P = 2.5 atm

1 atm = 101.300 J/L

initial volume Vi = 2.50 L

final  volume Vf = 4.50 L

change in volume ΔV = 2 L

The change in internal energy ΔU = ??

Solution

as we know that

U = q + w

for adiabatic process

q = 0

U = w

ΔU = - P ΔV

putting values

ΔU = - 2.5 X 2.0

ΔU = - 5  L X 101.3 J/L                            as 1 atm = 101.300 J/ L

ΔU = - 5 05.1 J

Answer 2

Answer:

Explanation:

Solution :

Work done in irreversible process

W=0PextΔV

=−2.5[4.5−2.5]=−5Latm

=−5×101.3J=−505J

Since system is well insulated q=0

From I law of thermodynamics ΔE=q+W

ΔE=W=−505J

Explanation: