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

Answer:

a) w= -95.53 KJ

b)Q= -95.53

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  

$W=P_1V_1\ ln\dfrac{V_2}{V_1}$   or

$W=mRT\ ln\dfrac{P_1}{P_2}$

For air$R=0.287\frac{J}{Kg-K}$  

T is temperature in Kelvin.

$W=0.287\times 303\ ln\dfrac{100}{300}$ 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.