Suppose 1.00Lof a gas with$\gamma =1.30$, initially at 273K and1.00atm is suddenly compressed adiabatically to half its initial volume. Find

1. Its final pressure
2. Its final temperature
3. If the gas is then cooled to 273Kat constant pressure, what is its final volume?

• Initial volume is$V_i=1.0L$
• Final volume is$V_f=0.5\hspace{0.33em}L$
• Initial pressure is$p_i=1.00a\mathrm{tm}$
• Temperature is T_i=273K
• Gamma isrole="math" localid="1662528092824" $\gamma =1.30$

In case of adiabatic process,

$p{V}^{\gamma }=Constant$

Here p is the pressure, V is the volume and$\gamma$ is the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

$\gamma =\frac{C_P}{C_V}$

Here C_Pis specific heat capacity at constant pressure and C_Vis specific heat capacity at constant volume.

We can write two equations,

$p_i{V}_i=nRT$....... (1)

role="math" localid="1662528411298" $p_f{V}_f=nRT$....... (2)

Dividing equation (2) by (1) we get,

$\frac{\left(p_f{V}_f\right)}{\left(p_i{V}_i\right)}=1$

We can write,

$p_i{V}_i^{\gamma }=p_f{V}_f^{\gamma }$

$\Rightarrow \frac{\left(V_i^{\gamma }\right)}{\left(V_f^{\gamma }\right)}=\frac{p_f}{p_i}$

$\Rightarrow p_f=p_i{\left(\frac{V_i}{V_f}\right)}^{\gamma }$

We are given that final volume is half of initial volume.

$V_f=0.5V_i$

We can write,

$p_f=p_i{\left(\frac{V_i}{0.5V_i}\right)}^{\gamma }$

role="math" localid="1662528896486" $p_f=1.00{\left(2\right)}^{1.30}=2.46a\mathrm{tm}$

Therefore the final pressure is 2.46atm.

From equation 19-56 we can write,

$T_f={\left(\frac{V_f}{V_i}\right)}^{\gamma -1}{T}_i$

Substitute 2V_ifor V_f,1.30 for$\gamma$ and 273K for T_iinto the above equation,

$\begin{array}{l}{T}_f=2^{1.30-1}\times 273\\ =336.1\\ =336K\end{array}$

Therefore the final temperature is 336K .

At constant pressure we can write,

$\frac{V_f}{V_i}=\frac{T_f}{T_i}$

Substitute 336k for T_i, 273K for T_f, 0.5L for V_i into the above equation,

$$\begin{gathered} V_f=\left(\frac{T_f}{T_i}\right)V_i \\ =\frac{273}{336}\left(0.5\right) \\ =0.046L \end{gathered}$$

Therefore the final volume is 0.046L.