91Ó°ÊÓ

• Initial volume is$V_i=4.3L$
• Final volume is $V_f=0.76L$
• Initial pressure is $p_i=1.2atm$
• Temperature is T_i= 310K
• $\mathrm{γ}=1.4$
  

In case of adiabatic process,

$p{V}^{\mathrm{γ}}=Cons\tan t$

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.

localid="1662398424168" $\mathrm{γ}=\frac{C_P}{C_V}$

Here C_P is specific heat capacity at constant pressure and C_Vis specific heat capacity at constant volume.

For adiabatic process, we can write

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

role="math" localid="1662442278944" $⇒\frac{p_i{V}_i^{\mathrm{γ}}}{V_f^{\mathrm{γ}}}=p_f$

Substitute 1.2 atm for p_i , 4.3L for V_i , 0.76L for V_finto the above equation,

$\begin{array}{l}{p}_f=1.2{\left[\frac{4.3}{0.76}\right]}^{1.4}\\ =13.5799atm\\ =14atm\end{array}$

Therefore the final pressure is 14atm .

We can write two equations according to the gas law as

Dividing second equation by the first, we get

$\frac{p_f{V}_f}{p_i{V}_i}=\frac{nR{T}_f}{nR{T}_i}$

Substitute 13.5799 atm for p_f , 0.76L for V_f ,1.2atm for p_i, 4.3L for V_i and 310K for T_I into the above equation,

$\begin{array}{l}{T}_{f=\frac{\left(13.5799×0.76\right)}{\left(1.2×4.3\right)}×310}\\ =6.2×{10}^{2}K\end{array}$

Therefore the final temperature is $6.2×{10}^{2}K$.