Question

The minimum work which must be done to
compress 16 gm of oxygen at 300 K from a
pressure of 1.01325 * 103 N/m2 to 1.01325 x
105 N/m2 is:
(A) 5744 J
(B) 8622 J
(C) 3872J
(D) 7963 J​

Answer 1

Answer:

5744 J

Explanation:

Given that = P1 = 1.01325 * 10^3 N/m2

P2 = 1.01325 *10^5 N/m2

T = 300 K

weight of oxygen = 16 g

number of moles = given mass/atomic mass ⇒ 16/32 = 0.5 mol

Minimum work done = -nrT㏑(P1/P2)

putting the value we get

= -0.5×8.314×300×㏑(10^3/10^5)

=W minimum  ⇒5744J

therefore answer = 5744 J