# Thermodynamics, first law, pressure increase, volume increase

## Homework Statement

How much heat energy is needed to heat oxygen, mass=10g which is on temperature t=27°C to make the volume go up by 3 times while pressure is constant, and after that increase the pressure 2 times while volume is constant. Specific heat capacity of oxygen is cp= 908 J/KgK and cv=653 J/kgK.

## Homework Equations

Q=m*cv(p)*delta(T)
Q=U+pV

## The Attempt at a Solution

I tried and tried and triiied and i hate to come here to look for an answer its a probably an easy one, but i have a mental blockade, can't think of anything.

I have final result Q=11,32kJ if it helps(solution to this from my book)

I simply cannot find any relation or equation so that i can relate pressure and volume etc. I know general fomula Q=U+pV, but that didn't get me anywhere..

Thanks

Last edited:

## Answers and Replies

pV/T = const

const pressure:
1/(27+273) = 3 / (T + 273) --> dT = 600
Q1 = mc_v dT = 0.01kg * 908 * 600 = 5448 KJ

const vol:
1 / (627 + 273) = 2 / (T + 273) --> dT = 900
Q2 = mc_pdT = 0.01kg * 653 * 900 = 5877 kJ

Q_total = Q1 + Q2 = 11325kJ

pV/T = const

const pressure:
1/(27+273) = 3 / (T + 273) --> dT = 600
Q1 = mc_v dT = 0.01kg * 908 * 600 = 5448 KJ

const vol:
1 / (627 + 273) = 2 / (T + 273) --> dT = 900
Q2 = mc_pdT = 0.01kg * 653 * 900 = 5877 kJ

Q_total = Q1 + Q2 = 11325kJ

See, I had absolutely no idea about pV/T part, that changes everything. THANK YOU VERY MUCH!!!!!!!!!!!!!!!!!!!!!!!!

I just had like 5 in a row "aha" moments :D Gas equation, fixed quantity of gas etc

Andrew Mason
Science Advisor
Homework Helper
pV/T = const

const pressure:
1/(27+273) = 3 / (T + 273) --> dT = 600
Q1 = mc_v dT = 0.01kg * 908 * 600 = 5448 KJ

const vol:
1 / (627 + 273) = 2 / (T + 273) --> dT = 900
Q2 = mc_pdT = 0.01kg * 653 * 900 = 5877 kJ

Q_total = Q1 + Q2 = 11325kJ
I think you meant to write C_p in the first calculation (Q1) and C_v in the second (Q2). You have used the correct values though.

AM