Question

A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m3m3. A constant pressure process gives 18 kJ of work out. What is the final temperature? You may assume ideal gas.

Answer 1

1160 K.

Step-by-step explanation:

Given that

Initial

Pressure P =600 KPa

Temperature T =290 K

Volume V =0.01
`m^3`

If we assume that air is s ideal gas the

P V = mRT

R=0.287 KJ/kg.k

now by putting the values in above equation

600 x 0.01 = m x 0.287 x 290

m=0.07 kg

The work out at constant pressure given as

`w=P(V_2-V_1)`

`18=600(V_2-0.01)`

`V_2=0.04\ m^3`

At constant pressure

`(T_2)/(T_1)=(V_2)/(V_1)`

`(T_2)/(290)=(0.04)/(0.01)`

`T_2=1160\ K`

So the final temperature is 1160 K.