Question

Air is compressed (in a piston cylinder device) isothermically from 100kPa and 30°C to 300kPa. On a per unit basis, compute: a) The amount of work that flows into the air. b) The change in energy of the air. c) The amount of heat that flows out of the air.

Answer 1

a) w= -95.53 KJ

b)Q= -95.53

Step-by-step explanation:

At initial condition
`P_1=100 KP,T_1=30°C`

At final condition
`P_2=300 KPa`

a)

It is constant temperature process so PV=C

`P_1V_1=P_2V_2`

So work in constant temperature process

`W=P_1V_1\ ln(V_2)/(V_1)` or

`W=mRT\ ln(P_1)/(P_2)`

For air
`R=0.287(J)/(Kg-K)`

T is temperature in Kelvin.

`W=0.287* 303\ ln(100)/(300)` KJ

So w= -95.53 KJ ,This is work for unit mass.

Negative sign indicates , it is compression process.

b)

If take air as ideal gas then We know that internal energy for ideal gas is a function of temperature only.Here process is a constant temperature process so temperature of air will remains constant so ,change internal energy of air is equal to zero.

`u_2-u_1=0`

c)

From first law of thermodynamics

Q=
`u_2-u_1`+W

Here
`u_2-u_1=0`

So Q=W ( w= -95.53 KJ )

⇒Q= -95.53 KJ

Negative sign indicates that heat flows out of the air.