4. charles’ law\n$\frac{v_1}{t_1}=\frac{v_2}{t_2}$\n$t_1 = 273 k$\n$v_1 = 2.45 l$\n$t_2 = 373 k$\n$v_2 =…

4. charles’ law\n$\frac{v_1}{t_1}=\frac{v_2}{t_2}$\n$t_1 = 273 k$\n$v_1 = 2.45 l$\n$t_2 = 373 k$\n$v_2 = $\n$t_1 = $\n$v_1 = 1.38 l$\n$t_2 = 86.0 k$\n$v_2 = 4.44 l$

4. charles’ law\n$\frac{v_1}{t_1}=\frac{v_2}{t_2}$\n$t_1 = 273 k$\n$v_1 = 2.45 l$\n$t_2 = 373 k$\n$v_2 = $\n$t_1 = $\n$v_1 = 1.38 l$\n$t_2 = 86.0 k$\n$v_2 = 4.44 l$

Answer

Explanation:

Step1: Use Charles' Law formula for first - row calculation

Given $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, we can solve for $V_2$. Rearranging the formula gives $V_2=\frac{V_1T_2}{T_1}$. Substitute $V_1 = 2.45\ L$, $T_1=273\ K$, and $T_2 = 373\ K$ into the formula. $V_2=\frac{2.45\times373}{273}$

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

$V_2=\frac{2.45\times373}{273}=\frac{913.85}{273}\approx3.35\ L$

Step3: Use Charles' Law formula for second - row calculation

Given $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, we can solve for $T_1$. Rearranging the formula gives $T_1=\frac{V_1T_2}{V_2}$. Substitute $V_1 = 1.38\ L$, $V_2 = 4.44\ L$, and $T_2 = 86.0\ K$ into the formula. $T_1=\frac{1.38\times86.0}{4.44}$

Step4: Calculate the value of $T_1$ for the second - row

$T_1=\frac{1.38\times86.0}{4.44}=\frac{118.68}{4.44}=26.7\ K$

Answer:

First - row $V_2\approx3.35\ L$, Second - row $T_1 = 26.7\ K$