Please help me to evaluate my answerPart $4-$ A gas is stored in a cylinder that has a volume of $V1$ litres. The gas is stored at temperature T$1$ and pressure $P1.$ The gas is then released into the air at sea level. Calculate the final temperature of the gas and the polytropic process index value $(n)$ of the gas.

$P_1=3$MPa

$T_1=350K$

$R=8,314\frac{K}{m}de*K$

$P_2=100$kPa

$V_m=22,4\frac{?}{?}\%$mole

$V_1=2,5l$

$V=V_m*n$

$P_1*V_1=n*R*T_1$

$V_m=\frac{V}{n}=\frac{R*T}{P}$

$n=\frac{P_1*V_1}{R*T_1}=\frac{3000\text{k}\text{P}\text{a}*2,5l}{8,314\frac{K}{m}{l}^K*350(K)}=2,577$

$n=2,577$

$V_m=\frac{V}{n}=V_2=V_m*n=22,5l$ role $*2,5=57,982l=V_2$

$V_m=\frac{R*T_2}{P_2}=>T_2=\frac{V_m*P_2}{R}=\frac{22,4\frac{l}{m}\text{o}\text{l}\text{e}*100\text{k}\text{P}\text{a}}{8,314\frac{K}{m}de*K}=269,42K$

$T_2=269,42K$

$p_1*V_1^{n1}=P_2*V_2^{n_1}$

$\frac{P_1}{P_2}=\frac{V_2^{n_1}}{V_1^{n_1}}$

$ln{P}_1-ln{P}_2=n_1*ln{V}_2-n_1*ln{V}_1$

$ln{P}_1-ln{P}_2=n_1*(ln{V}_2-ln{V}_1)$

$\frac{ln{P}_1-ln{P}_2}{ln{V}_2-ln{V}_1}=n_1=\frac{ln3000\text{k}\text{P}\text{a}-ln100\text{k}\text{P}\text{a}}{ln57,982l-ln2,5l}=1,08=n$