Question

2.
The maximum work done in expanding 16g oxygen at 300K and occupying a volume of 5dm’ isothermally
until the volume become 25dm’ is:
a) 2.01x10')
b) +2.81x10ºJ
c) 2.01x10'J
d) +2.01x10J​

Answer 1

Answer:

the maximum work done in expending 16g of oxygen at 300k and 0ccuping a volume of 5dm3 isothermally the volume becomes 25dm3 is in5=o.

Answer 2

Explanation:

work done = -2.303 × nRT × log ( V2 / V1 )

value of RT = 8.314 × 300

the value of ( n = no. of moles)

formula of n = ( amount of O2 / molecular mass of O2 )

therefore,

n = 16 / 32 = 0.5

then, value of

V1 = 5 dm*3

and V2 = 25 dm*3

the value puting the equation.

W = -2.303 × nRT × log ( V2 / V1 )

W = -2.303 × 0.5 × 8.314 × 300 × log ( 25 / 5 )

therefore ans is 2.01 × 10^3 Joule

the, W = 2.01 × 10^3 joule

i hope this is helpful to you..!!