Question

1 mole of an ideal gas at 300 K and 10 atm is expanded to 1.0 atm pressure isothermally
and reversibly then work involved is
5.74 kJ
5744 kJ
57.44 kJ
O 5.7 J
0

Answer 1

Explanation:

δS

surr

=

T

δq

Now δq=δu+δw

Since isothermal process

δu=0

Hence

δq=δw

Now,

Heat gained by the surrounding is heat lost by the system.

Hence

δq

system

=−δq

surr

Therefore,

δS

surr

=

T

δq

surr

=

T

−δq

sys

=

T

−δw

=

T

−P(dV)

=

300

−3(2−1)

×103 Joules

=−1.03 joules.

Answer 2

Answer:

Here is your answer mate