Question

1. gay lussac’s law: $\frac{p_1}{t_1}=\frac{p_2}{t_2}$

$p_1 = 3.40$ atm	$t_1 = 298$ k	$p_2 = 8.80$ atm	$t_2 =$
$p_1 =$	$t_1 = 100$ k	$p_2 = 6.62$ mmhg	$t_2 = 178$ k

Explanation:

Step1: Solve for $T_2$ in the first - set of values

Given $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we can cross - multiply to get $P_1T_2 = P_2T_1$. Then $T_2=\frac{P_2T_1}{P_1}$. Substitute $P_1 = 3.40$ atm, $T_1 = 298$ K, and $P_2 = 8.80$ atm into the formula:
$T_2=\frac{8.80\ atm\times298\ K}{3.40\ atm}$

Step2: Calculate the value of $T_2$

$T_2=\frac{8.80\times298}{3.40}=\frac{2622.4}{3.40}\approx771.3$ K

Step3: Solve for $P_1$ in the second - set of values

From $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we can cross - multiply to get $P_1T_2 = P_2T_1$. Then $P_1=\frac{P_2T_1}{T_2}$. Substitute $P_2 = 6.62$ mmHg, $T_1 = 100$ K, and $T_2 = 178$ K into the formula:
$P_1=\frac{6.62\ mmHg\times100\ K}{178\ K}$

Step4: Calculate the value of $P_1$

$P_1=\frac{662}{178}\approx3.72$ mmHg

Answer:

For the first part, $T_2\approx771.3$ K; for the second part, $P_1\approx3.72$ mmHg