Question

on4Suppose that the pressure of 1.23 L of gas is 300.6 mm Hg when the temperature is 218.5 K. At what temperature is the volume 8.32 L and the pressure 802.75 mm Hg?O a. 3950O b. 8610O c. 1250O d. 12.1Check

Answer 1

Answer:

The final temperature is 3,952K. (The closest option is 3,950K).

Explanation:

The given information from the exercise is:

- Initial pressure (P1): 300.6mmHg

- Initial volume (V1): 1.23L

- Initial temperature (T1): 218.5K

- Final volume (V2): 8.32L

- Final pressure (P2): 802.75mmHg

We can calculate the final temperature (T2), by replacing the values of P1, V1, T1, V2 and P2 in the following formula of Ideal Gases:

$\begin{gathered} \frac{P_1*V_1}{T_1}=\frac{P_2*V_2}{T_2} \\ \frac{300.6mmHg*1.23L}{218.5K}=\frac{802.75mmHg*8.32L}{T_2} \\ 1.69\frac{mmHgL}{K}=\frac{6,678.88mmHgL}{T_2} \\ T_2=\frac{6,678.88mmHgL}{1.69\frac{mmHgL}{K}} \\ T_2=3,952K \end{gathered}$

So, the final temperature is 3,952K. (The closest option is 3,950K).