- #1

Lewishio

- 19

- 2

- Homework Statement
- I have had a go at the following question but have gone wrong somewhere, any ideas?

A mass of 0.12 kg of air has an initial temperature of 500°C and pressure 0.8 MPa. If the air is

expanded according to the law pV1.2 = c to a final volume of 90 litres, determine

i) its initial volume,

ii) its final pressure,

iii) its final temperature.

For air, take R = 287 Jkgˉ¹ Kˉ¹.

- Relevant Equations
- pV = mRT

V = mRT/P

T = PV/mR

Mass = 0.12kg

Initial temp = 500°c = 773K

Initial pressure = 0.8 MPa = 800,000 Pa

Final volume = 90L

R = 287 Jkg^-1K^-1

1) Initial Volume

V=mRT/P

0.12 x 287 x 773 / 800,000 = 26,662.12m^3

2) Final Pressure

P2 = P1P2^1.2/V2^1.2

800,000 x 26,662.12^1.2 / 90 = 1,816,095,330 Pa = 1,816 MPa

3) Final Temperature

T=P2V2^1.2/mR

1,816,095,330 x 90 / 0.12 x 287 = 1,291 x 10^11 K

Initial temp = 500°c = 773K

Initial pressure = 0.8 MPa = 800,000 Pa

Final volume = 90L

R = 287 Jkg^-1K^-1

1) Initial Volume

V=mRT/P

0.12 x 287 x 773 / 800,000 = 26,662.12m^3

2) Final Pressure

P2 = P1P2^1.2/V2^1.2

800,000 x 26,662.12^1.2 / 90 = 1,816,095,330 Pa = 1,816 MPa

3) Final Temperature

T=P2V2^1.2/mR

1,816,095,330 x 90 / 0.12 x 287 = 1,291 x 10^11 K