Thermodynamics: Gas pressure and change in volume

In summary, the conversation discusses a thermodynamics question involving a gas held in a rigid container and heated at constant pressure. The specific heat capacities of the gas at constant pressure and constant volume are calculated and used to find the initial pressure and change in volume of the gas. The correct value for R is determined to be 0.5kJ/kg K, and the use of Boyle's law is discussed as an alternative method for calculating the change in volume. The correct values for Cp and Cv are clarified to be 1.5kJ/kg K and 1kJ/kg K, respectively.
  • #1
Mingsliced
18
0

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!
 
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  • #2
Mingsliced said:
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
 
  • #3
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.
 
  • #4
Mingsliced said:
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
 
  • #5
Ah excellent, thanks very much!
 
  • #6
Cp=1.5 and Cv equals 1 yeah?
 
  • #7
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.
 

1. What is gas pressure in thermodynamics?

Gas pressure in thermodynamics is a measure of the force exerted by gas molecules on the walls of a container. It is caused by the collisions of gas molecules with the walls of the container and is typically measured in units of Pascals (Pa) or atmospheres (atm).

2. How does gas pressure change with temperature in thermodynamics?

In thermodynamics, gas pressure is directly proportional to temperature. This means that as the temperature of a gas increases, the gas molecules move faster and collide with the container walls more frequently, resulting in an increase in gas pressure. Likewise, as the temperature decreases, the gas pressure decreases.

3. What is the relationship between gas pressure and volume in thermodynamics?

According to Boyle's Law, in thermodynamics, the pressure of a gas is inversely proportional to its volume at a constant temperature. This means that as the volume of a gas decreases, the gas molecules collide with the container walls more frequently, resulting in an increase in gas pressure. Similarly, as the volume of a gas increases, the gas pressure decreases.

4. What is the formula for calculating gas pressure in thermodynamics?

The formula for calculating gas pressure in thermodynamics is P = nRT/V, where P is pressure, n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, and V is the volume of the gas. This formula is known as the Ideal Gas Law and is used to calculate the pressure of an ideal gas in a closed system.

5. How does gas pressure change during a change in volume in thermodynamics?

In thermodynamics, as the volume of a gas decreases, the gas molecules collide with the container walls more frequently, resulting in an increase in gas pressure. Similarly, as the volume of a gas increases, the gas pressure decreases. This change in gas pressure is directly related to the change in volume and can be calculated using the Ideal Gas Law.

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