Question

A cylinder with a movable piston contains $n$moles of gas at a temperature higher than that of the surrounding environment. An external force on the piston keeps the pressure constant while the gas cools as $\Delta T=(\Delta T{)}_0{e}^{-t/\tau }$, where $\Delta T$is the temperature difference between the gas and the environment, $(\Delta T{)}_0$is the initial temperature difference, and$\tau$ is the time constant.

a. Find an expression for the rate at which the environment does work on the gas. Recall that the rate of doing work is power.

b. What power is initially supplied by the environment if $0.15\mathrm{mol}$of gas are initially $12°\mathrm{C}$warmer than the surroundings and cool with a time constant of $60\mathrm{s}$?

Short answer

a) The expression for the rate at which the environment does work on the gas is $P=nR\frac{\tau }{t^2}(\Delta T{)}_0{e}^{-\frac{\tau }{t}}$.

b) The initial power is zero.

Step-by-step solution

Given Information (Part a)

Number of moles$=n$

Gas cools at $\Delta T=(\Delta T{)}_0{\mathrm{e}}^{-t/\tau }$

$∆T=$The temperature difference between the gas and the environment,

$∆T_{0=}$Initial temperature difference

$\tau =$Constant time

Explanation (Part a)

(a) We are given that

$\Delta T=(\Delta T{)}_0{e}^{-\frac{\tau }{t}}$

What we want to find is the rate at which work is done. For this we need to consider that the piston keeps the pressure constant, i.e. this is an isobaric process. We know that in an isobaric process the work is given by

$W=p\Delta V$

From the ideal gas equation, we have

$pV=nRT\Rightarrow V=\frac{nR}{p}T$

Considering that the pressure remains constant during an isobaric process, obviously the mole number and the ideal gas constant $R$, obtains us

$\Delta V=\frac{nR}{p}\Delta T$

It is now clear that to find the power, - the rate at which work is done,-we need to find the rate at which the volume changes (since the process is isobaric).

To find the rate at which the volume changes we'll use the relation we just derived, realizing that the only time-dependent part is the temperature difference $\Delta T$.

To find the time rate of change of the temperature difference we'll have to differentiate it with respect to time.

We have$\frac{d}{dt}\Delta T=\frac{d}{dt}\left((\Delta T{)}_0{e}^{-\frac{\tau }{t}}\right)$

A careful derivation would reveal

$\frac{d}{dt}\Delta T=(\Delta T{)}_0\frac{d}{dt}\left(e^{-\frac{\tau }{t}}\right)=(\Delta T{)}_0{e}^{-\frac{\tau }{t}}\frac{d}{dt}\left(-\frac{\tau }{t}\right)$

Finally,$\frac{d}{dt}\Delta T=\frac{\tau }{t^2}(\Delta T{)}_0{e}^{-\frac{\tau }{t}}$

Explanation (Part a)

This leaves us with

$\frac{d}{dt}\Delta V=\frac{nR}{p}\frac{\tau }{t^2}(\Delta T{)}_0{e}^{-\frac{\tau }{t}}$

The power will therefore be

$P=\frac{d}{dt}W=\frac{d}{dt}p\Delta V=p\frac{d}{dt}\Delta V$

Finally, our result is

$P=nR\frac{\tau }{t^2}(\Delta T{)}_0{e}^{-\frac{\tau }{t}}$

Final Answer (Part a)

Hence, the expression for the rate at which the environment does work on the gas is$P=nR\frac{T}{t^2}(\Delta T{)}_0{e}^{-\frac{T}{t}}$

Given Information (Part b) 

Number of moles$=n$

Gas cools at$\Delta T=(\Delta T{)}_0{\mathrm{e}}^{-t/T}$

$\Delta T=$The temperature difference between the gas and the environment,

$\Delta T_0=$Initial temperature difference

$T=$Constant time

Explanation (Part b) 

(b) In this numerical case, we just need substitute the mole number $n=0.15$, the initial temperature difference $\Delta T=12°\mathrm{C}$and the time constant, as well as the time; that is, we have provided an expression giving the power as a function of time. We can say numerically that

$P\left(t\right)=\frac{0.15·8.314·60}{t^2}·12·e^{-\frac{60}{t}}$

What we need to do, however, is something else: to find the power at $t=0$, as we can see, we cannot substitute $t=0$. As a result, we'll have to calculate the limit when the time goes to zero of the power, which results in

$\underset{t\to 0}{\lim }P\left(t\right)=0$

Therefore the initial power required is zero.

Explanation (Part b) 

In the figure below we provide a graph of the function $e^{-\frac{a}{x}}$(in blue) and its derivative (in red), where $a$is a constant. As we expected, the power will be zero in the beginning. It will also be zero after a long time, since the gas will reach thermal equilibrium with the environment.

Final Answer (Part b)

Hence, the power is initially supplied by the environment if$0.15$mol of gas are initially${12}^{\circ }C$warmer than the surroundings and cool with a time constant of$60s$is zero.