2. boyles law: $p_1v_1 = p_2v_2$; $p_1 = 2.20$ atm, $v_1 = 5.60$ l, $p_2 = 8.80$ atm, $v_2 = $; $p_1 = $…

2. boyles law: $p_1v_1 = p_2v_2$; $p_1 = 2.20$ atm, $v_1 = 5.60$ l, $p_2 = 8.80$ atm, $v_2 = $; $p_1 = $, $v_1 = 2.35$ l, $p_2 = 6.62$ mmhg, $v_2 = 15.0$ l

2. boyles law: $p_1v_1 = p_2v_2$; $p_1 = 2.20$ atm, $v_1 = 5.60$ l, $p_2 = 8.80$ atm, $v_2 = $; $p_1 = $, $v_1 = 2.35$ l, $p_2 = 6.62$ mmhg, $v_2 = 15.0$ l

Answer

Explanation:

Step1: Solve for $V_2$ in first - row

Given $P_1V_1 = P_2V_2$, we can re - arrange to $V_2=\frac{P_1V_1}{P_2}$. Substitute $P_1 = 2.20$ atm, $V_1 = 5.60$ L and $P_2 = 8.80$ atm. $V_2=\frac{2.20\times5.60}{8.80}$

Step2: Calculate the value of $V_2$ in first - row

$V_2=\frac{12.32}{8.80}=1.40$ L

Step3: Solve for $P_1$ in second - row

From $P_1V_1 = P_2V_2$, we re - arrange to $P_1=\frac{P_2V_2}{V_1}$. Substitute $V_1 = 2.35$ L, $P_2 = 6.62$ mmHg and $V_2 = 15.0$ L. $P_1=\frac{6.62\times15.0}{2.35}$

Step4: Calculate the value of $P_1$ in second - row

$P_1=\frac{99.3}{2.35}\approx42.3$ mmHg

Answer:

First - row: $V_2 = 1.40$ L Second - row: $P_1\approx42.3$ mmHg