Question

calculate the maximum work in

24 grams of O2 are expanded isothermally and reversibly from the pressure of 1.6 bar

to 1 bar at 298 Kelvin​

Answer 1

Answer:

Solution:-

Given weight

=

24

g

=24g

Molecular weight

=

32

g

=32g

No. of moles

(

n

)

=

24

32

=

0.75

mol

(n)=2432=0.75 mol

As we know that work done in an isothermal process is given as-

W

=

−

2.303

n

R

T

log

P

1

P

2

W=−2.303nRTlog⁡P1P2

Given:-

T

=

298

K

T=298K

P

1

=

1.6

×

10

5

P

a

=

160

k

P

a

P1=1.6×105Pa=160kPa

P

2

=

100

k

P

a

P2=100kPa

Therefore,

W

=

−

2.303

×

0.75

×

0.0821

×

298

×

log

160

100

W=−2.303×0.75×0.0821×298×log⁡160100

⇒

W

=

−

42.26

×

log

1.6

=

−

8.62

a

t

m

−

L

⇒W=−42.26×log⁡1.6=−8.62atm−L