A gas can expanding adiabatically might stop expanding either because it has filled arigid container (known final volume) or it has equilibrated with an external pressure(known final volume).Derive the expression for the final pressure of an ideal gas after an adiabatic expansion.Begin with the simple expression derived using the ideal gas law:PiVi TiPf Vf TfThen use the expression for the ratio of the final temperatureR/C₂The final expression isTiTf==(1)Pf = Pi- (²*

Answered: A gas can expanding adiabatically might…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

A gas can expanding adiabatically might stop expanding either because it has filled a
rigid container (known final volume) or it has equilibrated with an external pressure
(known final volume).
Derive the expression for the final pressure of an ideal gas after an adiabatic expansion.
Begin with the simple expression derived using the ideal gas law:
PiVi Ti
Pf Vf Tf
Then use the expression for the ratio of the final temperature
R/C₂
The final expression is
Ti
Tf
=
=
(1)
Pf = Pi
- (²
*

Expert Solution

Step 1

According to 1st law of thermodynamics, dQ = dU + pdV ....... (1)

dQ is the heat absorbed by the system

dU is the change in internal energy

pdV is the mechanical work

C_p and C_v- heat capacity at constant pressure and volume respectively

Considering an ideal gas is expanded in a reversible adiabatic process from P₁, V₁, T₁ to P₂, V₂, T₂. P₁, V₁, T₁ are the initial pressure, volume and temperature respectively and P₂, V₂, T₂ are final pressure, volume and temperature respectively.

For an adiabatic process dQ = 0

then from eq 1

dU = -pdV

$∵ d U = n C_{v} d T + {(\frac{\partial U}{\partial V})}_{T} F o r i d e a l g a s {(\frac{\partial U}{\partial V})}_{T} = 0 S o d U = n C_{v} d T$

So, nC_vdT = -pdV

$n C_{V} d T = - (\frac{n R T}{V}) d V (\frac{C_{V}}{R}) \int_{T_{1}}^{T_{2}} \frac{d T}{T} = - \int_{V_{1}}^{V_{2}} \frac{d V}{V} (\frac{C_{V}}{R}) \ln \frac{T_{2}}{T_{1}} = \ln \frac{V_{1}}{V_{2}} \ln \frac{T_{2}}{T_{1}} = {(\ln \frac{V_{1}}{V_{2}})}^{\frac{R}{C_{V}}}$

$\ln \frac{T_{2}}{T_{1}} = {(\ln \frac{V_{1}}{V_{2}})}^{\frac{C_{P} - C_{V}}{C_{V}} (C_{P} - C_{V} = R)} \ln \frac{T_{2}}{T_{1}} = {(\ln \frac{V_{1}}{V_{2}})}^{γ - 1} γ = \frac{C_{P}}{C_{V}} T_{2} {(V_{2})}^{γ - 1} = T_{1} {(V_{1})}^{γ - 1} \frac{P_{2} V_{2}}{n R} {(V_{2})}^{γ - 1} = \frac{P_{1} V_{1}}{n R} {(V_{1})}^{γ - 1} P_{2} {(V_{2})}^{γ} = P_{1} {(V_{1})}^{γ} P_{2} = P_{1} \times {(\frac{V_{1}}{V_{2}})}^{γ} . . . . . . (2)$

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions