Questions: A sample of gas in a flexible container initially occupies a volume of 3.87 L at 1.47 atm. Then, the volume of the gas inside the container decreases to 3.45 L at 342 K and 1.87 atm. The number of moles of the gas remains constant.

A sample of gas in a flexible container initially occupies a volume of 3.87 L at 1.47 atm. Then, the volume of the gas inside the container decreases to 3.45 L at 342 K and 1.87 atm. The number of moles of the gas remains constant.
Transcript text: A sample of gas in a flexible container initially occupies a volume of 3.87 L at 1.47 atm. Then, the volume of the gas inside the container decreases to 3.45 L at 342 K and 1.87 atm. The number of moles of the gas remains constant.
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Solution

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Solution Steps

Step 1: Identify Given Values

We are given the following initial and final conditions for the gas:

  • Initial pressure, \( P_1 = 1.47 \) atm
  • Initial volume, \( V_1 = 3.87 \) L
  • Final pressure, \( P_2 = 1.87 \) atm
  • Final volume, \( V_2 = 3.45 \) L
  • Final temperature, \( T_2 = 342 \) K
Step 2: Use the Ideal Gas Law

Since the number of moles of gas remains constant, we can use the combined gas law: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Step 3: Solve for Initial Temperature \( T_1 \)

Rearrange the combined gas law to solve for \( T_1 \): \[ T_1 = \frac{P_1 V_1 T_2}{P_2 V_2} \]

Step 4: Substitute the Given Values

Substitute the given values into the equation: \[ T_1 = \frac{(1.47 \, \text{atm}) (3.87 \, \text{L}) (342 \, \text{K})}{(1.87 \, \text{atm}) (3.45 \, \text{L})} \]

Step 5: Calculate \( T_1 \)

Perform the calculation: \[ T_1 = \frac{1.47 \times 3.87 \times 342}{1.87 \times 3.45} = \frac{1945.146}{6.4615} \approx 301.0 \, \text{K} \]

Final Answer

\[ \boxed{T_1 = 301.0 \, \text{K}} \]

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