Question

1. a gas occupies 1.00 l at standard temperature. what is the volume at 333.0°c? 1 $\frac{606(x)606x}{273}$ $n_1$ $t_1 = 606$ $v_2?$ $t_2 = 273$

Explanation:

Step1: Identify the initial and final conditions

Initial volume $V_1 = 1.00\ L$, initial temperature $T_1=273\ K$ (standard temperature), final temperature $T_2 = 333.0 + 273=606\ K$ (convert Celsius to Kelvin by adding 273).

Step2: Apply Charles's Law

Charles's Law is $\frac{V_1}{T_1}=\frac{V_2}{T_2}$. We want to find $V_2$. Rearranging the formula gives $V_2=\frac{V_1T_2}{T_1}$.

Step3: Substitute the values

Substitute $V_1 = 1.00\ L$, $T_1 = 273\ K$, and $T_2=606\ K$ into the formula: $V_2=\frac{1.00\times606}{273}$.

Step4: Calculate the result

$V_2 = 2.22\ L$.

Answer:

$2.22\ L$