Question

1. using the gay-lussac formula for gases, solve the following problem: a closed rigid container measures 0.67atm 25 degrees celsius. what will the pressure be at 400 degrees celsius? $\frac{p_1}{t_1} = \frac{p_2}{t_2}$ remember to convert all degrees into kelvin +273.15. answer: _______

Explanation:

Step1: Convert temperatures to Kelvin

First, convert the initial temperature \( T_1 = 25^\circ\text{C} \) to Kelvin. Using the formula \( T(\text{Kelvin}) = T(\text{Celsius}) + 273.15 \), we get \( T_1 = 25 + 273.15 = 298.15\,\text{K} \).

Next, convert the final temperature \( T_2 = 400^\circ\text{C} \) to Kelvin: \( T_2 = 400 + 273.15 = 673.15\,\text{K} \).

Step2: Rearrange the Gay-Lussac formula to solve for \( P_2 \)

The Gay-Lussac formula is \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Rearranging for \( P_2 \), we get \( P_2 = P_1 \times \frac{T_2}{T_1} \).

Step3: Substitute the known values into the formula

We know \( P_1 = 0.67\,\text{atm} \), \( T_1 = 298.15\,\text{K} \), and \( T_2 = 673.15\,\text{K} \). Substituting these values in, we have:

\( P_2 = 0.67\,\text{atm} \times \frac{673.15\,\text{K}}{298.15\,\text{K}} \)

First, calculate the ratio \( \frac{673.15}{298.15} \approx 2.257 \).

Then, multiply by \( P_1 \): \( P_2 = 0.67 \times 2.257 \approx 1.512\,\text{atm} \).

Answer:

The pressure at 400 degrees Celsius will be approximately \( 1.51\,\text{atm} \) (or more precisely \( 1.512\,\text{atm} \)).