A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process to a final pressure of 4 bar, during which the pressure–volume relationship is pV1.1 = constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, = 1 bar

= 4 bar

pV¹°¹ = constant

From superheated propane table, at  = 1 bar and  = 27⁰C

= 0.557 m³/kg

= 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

At  = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

(b) Heat transfer = ?

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