Chemistry, asked by Anonymous, 11 months ago

The minimum work which must be done to compress 16 gm of oxygen at 300k from a pressure of 1.01325×10^3 N/m^2 to 1.01325×10^5 N/m^2 is:
(A)5744 j (B)8622 j (C)3872 j
(D)7963 j

Answers

Answered by ramsmedicine
0

Answer:

Given that

m  

O  

2

​  

 

​  

=16g

P  

1

​  

=1.01325×10  

3

, P  

2

​  

=1.01325×10  

5

 

T=300K

moles of O  

2

​  

=  

32

16

​  

=0.5

w  

max

​  

=nRTln(  

P  

1

​  

 

P  

2

​  

 

​  

)

=0.5×8.3×300×ln(  

10  

3

 

10  

5

 

​  

)

=0.5×8.3×300×4.6

w  

max

​  

=5727J

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