91Ó°ÊÓ

Temperature$=300k$

Gas cylinder length$=10.0cm$

Spring is compressed $=2.0cm$

Gas cylinder length$=16.0cm$

Therefore, the first law of thermodynamics will be observed.

$\mathrm{Δ}E=W+Q⇒\underset{¯}{Q=\mathrm{Δ}E-W}$

A gas' work does the work of a spring. So, in this case, the gas is doing the spring's work, which, as we all know, is given by the spring's work.

$W=\frac{k{\left(x_2-x_1\right)}^2}{2}$

$x_1$and $x_2$are initial and final positions.

It is interesting to notice however that the difference in positions will be equal to the difference in length of the cylindrical chamber, as long as we assume that the piston is incompressible.

$W=\frac{k{\left(L_2-L_1\right)}^2}{2}$

Change in internal energy will be,

$\mathrm{Δ}E=n{C}_v\mathrm{Δ}T$

Number $n$of moles, as the piston moves from one length to another, there is a temperature difference in the gas $L_1=0.1\mathrm{m}$to $L_2=0.1m$

Ideal gas law applied at the initial state,

$p_1{V}_1=nR{T}_1⇒n=\frac{p_1{V}_1}{R{T}_1}$

In the prior section, we were not given the initial pressure, nor the initial volume, but, if we assume that the cross-sectional area of the piston is $A$, the initial volume is,

$V_1=A{L}_1$

Hence, the pressure is

$p_1=\frac{F_1}{A}=\frac{k{x}_0}{A},$

At equilibrium, the spring, which has a compression of $x_0=0.02\mathrm{m}$, is used to counterbalance this pressure coming from the gas.

$n=\frac{\frac{k{x}_0}{A}A{L}_1}{R{T}_1}=\frac{k{x}_0{L}_1}{R{T}_1}$

Calculate the difference in temperature,

$\mathrm{Δ}T=T_2-T_1=\left(\frac{T_2}{T_1}-1\right)T_1$

When the number of moles is constant and the ideal gas law is applied, one can derive the following result:

$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}⇒\frac{T_2}{T_1}=\frac{p_2}{p_1}\frac{V_2}{V_1}$

Assuming the volume to be a multiple of a length we can simplify to $V_x=A{L}_x$

$\frac{T_2}{T_1}=\frac{p_2}{p_1}\frac{L_2}{L_1}$

As a result, we need to calculate the ratio of the pressures again. For this we must remember the pressure after compression equals the initial pressure plus the extra pressure from the spring's "extra" compression. That is, the pressure after compression equals the initial pressure plus the extra pressure from the spring's "extra" compression over the area.

$p_2=p_1+\frac{F_2}{A}=p_1+\frac{k\left(x_2-x_1\right)}{2A}$

We have $x_2-x_1=L_2-L_1$.

By substituting this we get

$p_2=p_1+\frac{k\left(L_2-L_1\right)}{2A}=\frac{2A{p}_1+k\left(L_2-L_1\right)}{2A}$

Ratio of temperatures is,

$\frac{T_2}{T_1}=\frac{2A{p}_1+k\left(L_2-L_1\right)}{2A{p}_1}·\frac{L_2}{L_1}$

To change what we have been saying, we should replace $p_1=\frac{k{x}_0}{A}$with the simpler spring constant $k.$

$\frac{T_2}{T_1}=\frac{2x_0+\left(L_2-L_1\right)}{2x_0}·\frac{L_2}{L_1}$

We will now calculate the temperature difference based on the ratio of temperatures,

$\mathrm{Δ}T=\left(\frac{T_2}{T_1}-1\right)T_1=\left(\frac{2x_0+\left(L_2-L_1\right)}{2x_0}·\frac{L_2}{L_1}-1\right)T_1$

$\mathrm{Δ}T=\frac{2x_0{L}_2+L_2^2-L_1{L}_2-2x_0{L}_1}{2x_0{L}_1}{T}_1=\frac{L_2^2+2x_0\left(L_2-L_1\right)-L_1{L}_2}{2x_0{L}_1}{T}_1$

Change the expression in internal energy,

$\mathrm{Δ}E=n{C}_v\mathrm{Δ}T=\frac{k{x}_0{L}_1}{R{T}_1}{C}_v\frac{L_2^2+2x_0\left(L_2-L_1\right)-L_1{L}_2}{2x_0{L}_1}{T}_1$

$\mathrm{Δ}E=\frac{k{C}_v\left(L_2^2+2x_0\left(L_2-L_1\right)-L_1{L}_2\right)}{2R}$

We can combine the changes in energy and work and find the heat as the product of the combining,

$Q=\mathrm{Δ}E-W=\frac{k{C}_v\left(L_2^2+2x_0\left(L_2-L_1\right)-L_1{L}_2\right)}{2R}-\frac{k{\left(L_2-L_1\right)}^2}{2}$

On factoring,

$Q=\frac{k}{2R}\left\{C_v\left[L_2^2+2x_0\left(L_2-L_1\right)-L_1{L}_2\right]-R{\left(L_2-L_1\right)}^2\right\}$

The expression has no unknown parameters in it, so it can now be substituted numerically to find,

$Q=\frac{2000}{2·8.314}\left\{12.5·\left[0.{16}^2+2·0.02·(0.16-0.1)-0.1·0.16\right]-8.314·(0.16-0.1{)}^2\right\}$

$Q=14.4\mathrm{J}$

Hence, the amount of heat energy that must be added to the gas to expand the cylinder length$16.0cm$is$14.4J.$