Thermodynamics: Gas pressure and change in volume

[Mingsliced]

Homework Statement

Just want to check that I've used the correct method for this thermodynamics question I've been set. Any clarification would be greatly appreciated.

So I have 0.5kg of gas held in a rigid container of volume 0.25m^3 at a temperature of 20°C. 20kJ of heat energy is required to raise the temperature of the gas to 60°C.

When the same mass of gas is heated at a constant pressure, 30kJ of heat energy is required to create the same temperature rise.

I need to find:
ii) The initial pressure of the gas.
iii) The change in volume of the gas when it was heated at constant pressure.

I've already calculated the specific heat capacities of the gas at constant pressure and constant volume and believe this is correct:

R = (Cp = 1.5kJ/kG K) - (Cv = 0.75kJ/kG K)

2. The attempt at a solution

ii) V = 0.25m^3
T1 = 20°C (273 + 20 =293K)
T2 = 60°C (273 + 60 = 333K)
R = 0.75kJ/kG K
Q = 20kJ & 30kJ

Initial Pressure: PV = MRT

Transposed: P = MRT/V

P = 0.5 * (0.75*10^3) * 293/0.25

P = 439500 N/M^-2

iii) PV=MRT

Transposed: V = MRT/P

V = 0.5 * (0.75*10^3) * 333 / 439500

V = 124875 / 439500

V = 0.28m^3

Alternatively, I believe part iii) can be calculated with Boyle's Law (V1T1=V2T2), especially as the question says 'constant pressure'. This gives an answer of 0.22m^3. Not quite sure which would be the correct method.

Thanks for any help in advance!

 

[Andrew Mason]

So I have 0.5kg of gas held in a rigid container of volume 0.25m^3 at a temperature of 20°C. 20kJ of heat energy is required to raise the temperature of the gas to 60°C.

When the same mass of gas is heated at a constant pressure, 30kJ of heat energy is required to create the same temperature rise.

I need to find:
ii) The initial pressure of the gas.
iii) The change in volume of the gas when it was heated at constant pressure.

I've already calculated the specific heat capacities of the gas at constant pressure and constant volume and believe this is correct:

R = (Cp = 1.5kJ/kG K) - (Cv = 0.75kJ/kG K)

How do you get that value for Cv? It takes 20KJ of heat flow to raise the temperature of .5 kg by 40K. So it takes 40KJ to raise 1Kg by the same amount, or 1KJ/Kg K.

2. The attempt at a solution

ii) V = 0.25m^3
T1 = 20°C (273 + 20 =293K)
T2 = 60°C (273 + 60 = 333K)
R = 0.75kJ/kG K
Q = 20kJ & 30kJ

Initial Pressure: PV = MRT

Are we to assume the question states that this is an ideal gas?

Your method appears to be correct. Once you determine the value for R correctly, you should get the right answer.

AM

 

[Mingsliced]

Ah, I think I can see where I've gone wrong...

I was using Q = M * Cp (T2 - T1) and Q = Cv (T2 - T1), but using Q=30kj instead of 20kj.

So Cp = 1kj/kg K

and Cv = 0.5kj/kg K

Therefore R = 0.5kj/kg K

Is this correct? Thankyou :)

Yes, I believe that it is stated as an ideal gas.

 

[Andrew Mason]

Ah, I think I can see where I've gone wrong...

I was using Q = M * Cp (T2 - T1) and Q = Cv (T2 - T1), but using Q=30kj instead of 20kj.

So Cp = 1kj/kg K

and Cv = 0.5kj/kg K

Therefore R = 0.5kj/kg K

You got the right answer but your values for Cp and Cv are wrong.

$C_v = Q_v/M\Delta T = 20KJ/(.5Kg * 40K)$
$C_p = Q_p/M\Delta T = 30KJ/(.5Kg * 40K)$

AM

 

[Mingsliced]

Ah excellent, thanks very much!

 

[James FC]

Cp=1.5 and Cv equals 1 yeah?

 

[Chestermiller]

In your first post, you stated Boyle's law incorrectly. If you had stated it correctly, you would have gotten the final volumes to match.