Question

a gas at 300 k and 4.0 atm is moved to a new location with a temperature of 250 k. the volume changes from 5.5 l to 2.0 l. what is the pressure of the gas at the new location? use the formula: \\(\frac{p_1v_1}{t_1} = \frac{p_2v_2}{t_2}\\) 0.15 atm 2.3 atm 9.2 atm 14 atm

Explanation:

Step1: Identify known values

We know \( P_1 = 4.0 \, \text{atm} \), \( V_1 = 5.5 \, \text{L} \), \( T_1 = 300 \, \text{K} \), \( T_2 = 250 \, \text{K} \), \( V_2 = 2.0 \, \text{L} \), and the formula \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \). We need to solve for \( P_2 \).

Step2: Rearrange the formula for \( P_2 \)

Multiply both sides of the formula by \( T_2 \) and divide by \( V_2 \) to isolate \( P_2 \):
\( P_2 = \frac{P_1V_1T_2}{T_1V_2} \)

Step3: Substitute the known values

Substitute \( P_1 = 4.0 \), \( V_1 = 5.5 \), \( T_2 = 250 \), \( T_1 = 300 \), and \( V_2 = 2.0 \) into the formula:
\( P_2 = \frac{(4.0)(5.5)(250)}{(300)(2.0)} \)

Step4: Calculate the numerator and denominator

First, calculate the numerator: \( (4.0)(5.5)(250) = 4.0 \times 5.5 \times 250 = 22 \times 250 = 5500 \)
Then, calculate the denominator: \( (300)(2.0) = 600 \)

Step5: Divide to find \( P_2 \)

\( P_2 = \frac{5500}{600} \approx 9.2 \, \text{atm} \)

Answer:

9.2 atm