Pressure and temperature changes adiabatically for an ideal gas?

In summary, in this conversation, the question was asked to find the final pressure of an ideal gas that expands adiabatically to 1.51 times its initial volume. It was clarified that the gas is monatomic. Using the law for adiabatic expansion of an ideal gas and the values for Cv and Cp, which are 3/2R and 5/2R respectively, the ratio of Cp/Cv can be found to calculate gamma. This information was provided by mezarashi and Chi Meson.
  • #1
erik-the-red
89
1
Question:

An ideal gas, which is initially at a pressure of 4.05 atm and a temperature of 355 K is permitted to expand adiabatically to 1.51 times its initial volume.

A.
Find the final pressure if the gas is monatomic.

I was thinking [tex]P_i \cdot V_i = P_f \cdot V_f[/tex]. But, I made no use of the information that the gas is monatomic. Later on, a question asks for the final pressure if the gas is diatomic. Well, my starting point wouldn't distinguish between the two, so it's not right.
 
Physics news on Phys.org
  • #2
I would suppose you need to use the law for adiabatic expansion of an ideal gas. Of course, the ideal gas law holds as well.

[tex]PV^\gamma = constant[/tex]

Being monoatomic, the values for Cv and Cp are [tex]C_v = \frac{3}{2}R, C_p = C_v + R[/tex], which allows you to find gamma, as [tex]\gamma = \frac{C_p}{C_v}[/tex].
 
  • #3
erik-the-red said:
Question:
An ideal gas, which is initially at a pressure of 4.05 atm and a temperature of 355 K is permitted to expand adiabatically to 1.51 times its initial volume.
A.
Find the final pressure if the gas is monatomic.
I was thinking [tex]P_i \cdot V_i = P_f \cdot V_f[/tex]. But, I made no use of the information that the gas is monatomic. Later on, a question asks for the final pressure if the gas is diatomic. Well, my starting point wouldn't distinguish between the two, so it's not right.
The reason you can't use [tex]P_i \cdot V_i = P_f \cdot V_f[/tex] is becuase the temperature does not remain constant.
 
  • #4
Thanks mezarashi, that is what I needed.

Chi Meson, thanks for reminding me that temperature is not constant.
 
  • #5
Can someone explain this further? I don't understand what to use for gamma
 
  • #6
mezarashi had it right - gamma is the ratio of Cp / Cv.

Check your text for Cv of a monatomic ideal gas, and Cv of a diatomic ideal gas, then use the fact that Cp is Cv + R, for an ideal gas.
 

Related to Pressure and temperature changes adiabatically for an ideal gas?

1. What is an ideal gas?

An ideal gas is a theoretical gas that behaves according to the ideal gas law, which states that the pressure, volume, and temperature of a gas are related by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

2. What does it mean for a process to be adiabatic?

An adiabatic process is one in which there is no transfer of heat between the system and its surroundings. This means that the change in internal energy of the system is equal to the work done on or by the system.

3. How does pressure change in an adiabatic process for an ideal gas?

In an adiabatic process for an ideal gas, pressure and volume are inversely proportional. This means that as the volume of the gas decreases, the pressure increases, and vice versa.

4. How does temperature change in an adiabatic process for an ideal gas?

In an adiabatic process for an ideal gas, temperature and volume are directly proportional. This means that as the volume of the gas decreases, the temperature also decreases, and vice versa.

5. What is the equation for adiabatic temperature change in an ideal gas?

The equation for adiabatic temperature change in an ideal gas is T2/T1 = (V2/V1)^(gamma-1), where T2 and T1 are the final and initial temperatures, V2 and V1 are the final and initial volumes, and gamma is the adiabatic index, which is equal to the ratio of the specific heats of the gas.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
982
  • Introductory Physics Homework Help
Replies
2
Views
619
  • Introductory Physics Homework Help
Replies
1
Views
537
  • Introductory Physics Homework Help
Replies
33
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
810
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
723
Back
Top