A rigid tank containing 0.5kg of air was at 350 K and 1 atm. Heat is added to the air. And the final temperature becomes 500 K. Determine the entropy change of the air associated with theheating processin kJ/K.

the entropy change of the air associated with the heating process is 0.128 kJ/k

Explanation:

Given that;

m = 0.5 kg

T1 = 350 k

P1 = 1 atm

T2 = 500 k

since its a Rigid Tank, volume remains constant V1 = V2

SO

P1V1/T1 = P2V2/T2

P1/T1 = P2/T2

1 atm / 350 k = P2 / 500 k

350P2 = 500

P2 = 500 / 350

P2 = 1.428 atm

Now entropy change for a process is;

we know that; Cp = 1.005 and R = 0.287

ΔS = m[ Cpen T2/T1 - Renp2/p1 ]

= 0.5 [ 1.005 in( 500/350 ) - 0.287 in( 1.428/1 )

= 0.5 ( 0.3584 - 0.1022 )

= 0.5 × 0.2562      

ΔS = 0.128 kJ/k

Therefore the entropy change of the air associated with the heating process is 0.128 kJ/k

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

can you make it more obvious. i can't see it properly