91Ó°ÊÓ

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

In case of adiabatic process,

$p{V}^{\mathrm{γ}}=Constant$

Here p is the pressure, V is the volume and$\mathrm{γ}$ is the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

$\mathrm{γ}=\frac{C_P}{C_V}$

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,

$p_i{V}_i^{\mathrm{γ}}=p_f{V}_f^{\mathrm{γ}}$

$⇒\frac{\left(V_i^{\mathrm{γ}}\right)}{\left(V_f^{\mathrm{γ}}\right)}=\frac{p_f}{p_i}$

$⇒p_f=p_i{\left(\frac{V_i}{V_f}\right)}^{\mathrm{γ}}$

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

$V_f=0.5V_i$

We can write,

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

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,

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

Substitute 2V_ifor V_f,1.30 for$\mathrm{γ}$ and 273K for T_iinto the above equation,

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

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.