91Ó°ÊÓ

• Initial volume of the gas is V_i=200L
• Final volume of the gas is V_f=74.3L
• Initial pressure is$p_i=1.00atm$
• Final pressure is$p_f=4.00atm$

Initial temperature is T_I=300K

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.

For adiabatic process,

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

Hence,

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

Substitute 4.00atm for p_f ,1.00atm for p_i, 200Lfor V_Iand 74.3L for V_f into the above equation,

$\frac{4.00}{1.00}=\begin{array}{c}{\left(\frac{200}{743}\right)}^{\mathrm{γ}}\\ 4={(2.691)}^y\end{array}$

Taking log on both sides we get,

$\begin{array}{l}In\left(4\right)=\mathrm{γ}In(2.691)\\ \mathrm{γ}=\frac{In\left(4\right)}{In(2.691)}\\ =1.4\\ =\frac{7}{5}\end{array}$

From the table 19-3 we can conclude that it is a diatomic gas.

Ideal gas law equations for initial and final states are

$p_i{V}_i=nR{T}_i$…… (1)

$p_f{V}_f=nR{T}_f$…… (2)

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

localid="1663961348430" $\frac{p_f{V}_f}{p_i{V}_i}=\frac{nR{T}_f}{nR{T}_i}$

$\frac{p_f{V}_f}{p_i{V}_i}=\frac{T_f}{T_i}$

$T_f=T_i\frac{p_f{V}_f}{p_i{V}_i}$

Substitute 300K for T_i, 4atm for p_f, 74.3L for V_f , 1atm for p_i and 200L for V_i into the above equation,

localid="1662532760854" $\begin{array}{l}{T}_T=\frac{300(4×74.3)}{1×200}\\ =445.8\\ =446K\end{array}$

Therefore the final temperature is 446K .

Using ideal gas law we get,

$p_i{V}_i=nR{T}_i$$⇒n=\left(p_i{V}_i\right)/\left(R{T}_i\right)$

Substitute$1.01×{10}^{5}pa$ for p_i, 200L for V_i,$8.3145J/molâ‹\dots K$ for R and 300K for T_Iinto the above equation,

$$\begin{gathered} n=\frac{1.01×{10}^5×200}{8.314×300} \\ =8.10×{10}^{3}moles \end{gathered}$$

The number of moles in the gas is n= 8.10$×$10³ moles.