Question

a gas is contained in a thick - walled balloon under the conditions 2.95 atm, 379 k, and 7.456 l. the volume and temperature decrease to 4.782 l and 212 k, respectively. what is the final pressure of the gas in the balloon? pressure (atm)

Explanation:

Step1: Recall combined gas law

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

Step2: Rearrange for final pressure

$$P_2 = \frac{P_1V_1T_2}{T_1V_2}$$

Step3: Substitute values

$$P_2 = \frac{(2.95\ \text{atm})(7.456\ \text{L})(212\ \text{K})}{(379\ \text{K})(4.782\ \text{L})}$$

Step4: Calculate numerator

$$2.95 \times 7.456 \times 212 = 2.95 \times 1580.672 = 4662.9824$$

Step5: Calculate denominator

$$379 \times 4.782 = 1812.378$$

Step6: Compute final pressure

$$P_2 = \frac{4662.9824}{1812.378} \approx 2.57\ \text{atm}$$

Answer:

2.57 atm