Thermodynamics, first law, pressure increase, volume increase

[Bassalisk]

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

 

[ppzmis]

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

 

[Bassalisk]

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]

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

 

[paranormal]

for reaching out for help with this problem. First, let's define some of the variables in this problem. Q represents the amount of heat energy needed, m represents the mass of oxygen, cv represents the specific heat capacity at constant volume, and t represents the initial temperature. The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system. In this case, since the pressure is constant, there is no work done, so the equation simplifies to ΔU = Q. We can also use the equation Q = m*cv*ΔT to calculate the amount of heat energy needed to change the temperature of the oxygen.

For the first part of the problem, we need to find the amount of heat energy needed to increase the volume of the oxygen by a factor of 3 while keeping the pressure constant. We can use the ideal gas law, PV = nRT, to relate the initial and final volumes of the oxygen. Since the pressure is constant, we can rewrite this as V2 = V1*(T2/T1), where V1 is the initial volume and V2 is the final volume. We also know that T2 = T1 + ΔT, so we can substitute this into our equation to get V2 = V1*(1 + ΔT/T1). We also know that the final volume is 3 times the initial volume, so we can set up the equation 3V1 = V1*(1 + ΔT/T1) and solve for ΔT. This gives us ΔT = 2T1. Now we can use the equation Q = m*cv*ΔT to calculate the amount of heat energy needed, which gives us Q = 10g * 653 J/kgK * 2T1 = 13.06T1 J.

For the second part of the problem, we need to find the amount of heat energy needed to increase the pressure of the oxygen by a factor of 2 while keeping the volume constant. We can use the ideal gas law again, PV = nRT, to relate the initial and final pressures of the oxygen. Since the volume is constant, we can rewrite this as P2 = P1*(T2/T1), where P1 is the initial pressure and P2 is the final pressure. We also know that