CHALLENGE 7 A 30 g gas sample of CO₂ occupies 27 L and exerts a pressure of 5.6 atm. Additionally, the temperature measured is exactly 263 K within the container. The pressure is then tripled, the volume is reduced by its original value, and 25% of the gas leaks out. Calculate the new resulting temperature.

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### Challenge 7

A 30 g gas sample of CO₂ occupies 27 L and exerts a pressure of 5.6 atm. Additionally, the temperature measured is exactly 263 K within the container. The pressure is then tripled, the volume is reduced by ⅓ its original value, and 25% of the gas leaks out. Calculate the new resulting temperature.

**Detailed Explanation of the Problem:**
1. **Initial Conditions:**
   - Mass of gas \( m = 30 \) g
   - Volume \( V_1 = 27 \) L
   - Pressure \( P_1 = 5.6 \) atm
   - Temperature \( T_1 = 263 \) K

2. **Changes Applied to the Gas:**
   - Pressure is tripled: \( P_2 = 3 \times P_1 \)
   - Volume is reduced to ⅔ of its original value: \( V_2 = \frac{2}{3} V_1 \)
   - 25% of the gas leaks out, so 75% of the original gas remains.

**Hint:**
You may need to use the Ideal Gas Law and the relationships between pressure, volume, and temperature to solve for the new temperature after the changes have been applied.

#### Ideal Gas Law:
\[ PV = nRT \]
Where:
- \( P \) is the pressure
- \( V \) is the volume
- \( n \) is the number of moles
- \( R \) is the ideal gas constant
- \( T \) is the temperature in Kelvin

Use the initial conditions to find the number of moles \( n \) of the gas, and then apply the changes and the new conditions to calculate the resulting temperature \( T_2 \).
Transcribed Image Text:### Challenge 7 A 30 g gas sample of CO₂ occupies 27 L and exerts a pressure of 5.6 atm. Additionally, the temperature measured is exactly 263 K within the container. The pressure is then tripled, the volume is reduced by ⅓ its original value, and 25% of the gas leaks out. Calculate the new resulting temperature. **Detailed Explanation of the Problem:** 1. **Initial Conditions:** - Mass of gas \( m = 30 \) g - Volume \( V_1 = 27 \) L - Pressure \( P_1 = 5.6 \) atm - Temperature \( T_1 = 263 \) K 2. **Changes Applied to the Gas:** - Pressure is tripled: \( P_2 = 3 \times P_1 \) - Volume is reduced to ⅔ of its original value: \( V_2 = \frac{2}{3} V_1 \) - 25% of the gas leaks out, so 75% of the original gas remains. **Hint:** You may need to use the Ideal Gas Law and the relationships between pressure, volume, and temperature to solve for the new temperature after the changes have been applied. #### Ideal Gas Law: \[ PV = nRT \] Where: - \( P \) is the pressure - \( V \) is the volume - \( n \) is the number of moles - \( R \) is the ideal gas constant - \( T \) is the temperature in Kelvin Use the initial conditions to find the number of moles \( n \) of the gas, and then apply the changes and the new conditions to calculate the resulting temperature \( T_2 \).
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