The Setup
A Tale of Two Gases
Imagine you are an engineer tasked with compressing a very specific mixture of gases. Inside our perfectly insulated cylinder, we have a bustling crowd of particles: 5 moles of a monatomic gas (think of them as tiny, independent spheres zipping around) and 1 mole of a rigid diatomic gas (like tiny dumbbells tumbling through space).
The system starts at an initial pressure P0, volume V0, and temperature T0. Suddenly, we push the piston down, compressing the gas adiabatically until its volume shrinks to V0/4. Our mission is to uncover the final state of this mixture and the work required to achieve it.
Finding the Mixture's Identity
The Adiabatic Index
Before we can predict how the gas will behave under compression, we need to determine its thermodynamic identity. A mixture of gases behaves like a single, new gas with its own unique properties. The most crucial property here is the molar heat capacity at constant volume, (CV)mix.
We calculate this using a weighted average based on the number of moles:
(CV)mix=n1+n2n1CV1+n2CV2
For our monatomic gas,
CV1=23R, and for the diatomic gas,
CV2=25R. Substituting these values:
(CV)mix=5+15(23R)+1(25R)=6215R+25R=610R=35R
With
(CV)mix in hand, finding the adiabatic index (or ratio of specific heats),
γmix, is straightforward:
γmix=1+(CV)mixR=1+35RR=1+53=58=1.6
Boom! We've just validated Option (D). The adiabatic constant of the mixture is indeed 1.6.
The Pressure Cooker
Adiabatic Compression
Now, let's compress the gas. In an adiabatic process, no heat escapes or enters the system. The relationship between pressure and volume is governed by the beautiful equation:
PiViγ=PfVfγ
Let's plug in our initial and final states:
P0V01.6=Pf(4V0)1.6
Rearranging to solve for the final pressure
Pf:
Pf=P0(V0/4V0)1.6=P0(4)1.6
Here is where the problem throws us a mathematical lifeline. We know
4=22, so
41.6=(22)1.6=23.2. The problem explicitly states that
23.2=9.2.
Pf=9.2P0
This final pressure is clearly nestled between 9P0 and 10P0, making Option (A) absolutely correct.
The Heat is On
Finding the Final Temperature
As we compress the gas, we are doing work on it, which increases its internal energy and, consequently, its temperature. To find the final temperature Tf, we turn to our trusty ideal gas law: PV=nRT.
For the initial state, the total number of moles is
n=5+1=6. So,
P0V0=6RT0.
For the final state,
Tf=nRPfVf. Let's substitute what we know:
Tf=6R(9.2P0)(4V0)=49.2(6RP0V0)
Since
6RP0V0=T0, we get:
Tf=2.3T0
The gas has more than doubled in temperature! Now, let's check the "average kinetic energy" mentioned in Option (B). In this context, it refers to the total internal energy
U of the mixture.
U=n1CV1Tf+n2CV2Tf=(CV)mixnTf
U=(35R)(6)(2.3T0)=10R(2.3T0)=23RT0
Option (B) claims the energy is between 18RT0 and 19RT0. Our result of 23RT0 shatters this claim. Option (B) is incorrect.
Energy and Work
The Final Tally
Finally, let's calculate the work done during this intense compression. The formula for work in an adiabatic process is:
W=γ−1PiVi−PfVf
Substituting our knowns:
W=1.6−1P0V0−(9.2P0)(4V0)=0.6P0V0−2.3P0V0=0.6−1.3P0V0
To express this in terms of temperature, we recall that
P0V0=6RT0:
W=0.6−1.3(6RT0)=−13RT0
The negative sign perfectly aligns with our physical intuition: work is being done on the gas to compress it. The magnitude of the work, ∣W∣, is exactly 13RT0.
This confirms Option (C) is correct. We have successfully navigated the thermodynamics of this mixture, proving that options A, C, and D are the true statements!
