300k Initial temperature (T₁) = 200k. Use the combined gas law to solve the following problems: 8) If I initially have a gas at a pressure if 12 atm, a volume of 23 liters, and a temperature of 200.0 K, and then I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of the gas?V=Vita VP, VI Pava ta T₁ T₁. Pa 30. L atm x 23LX3000K X T

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Can you help me with the number 8 question? Can you explain step by step including the formula (use the combined gas law to solve)? I need to plug in the fraction to give the correct answer.
# Educational Text on Gas Laws

## Combined Gas Law Problems

### Example Problem 1:

**Problem Statement:**
If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

**Solution Steps:**
1. Use the formula: \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \)
2. Substitute known values:
   - \( P_1 = 12 \, \text{atm} \)
   - \( V_1 = 23 \, \text{L} \)
   - \( T_1 = 200 \, \text{K} \)
   - \( P_2 = 14 \, \text{atm} \)
   - \( T_2 = 300 \, \text{K} \)
3. Solve for \( V_2 \): 
   \[
   V_2 = \frac{(12 \, \text{atm}) \times (23 \, \text{L}) \times (300 \, \text{K})}{(14 \, \text{atm}) \times (200 \, \text{K})} = 29.57 \, \text{L}
   \]

### Example Problem 2:

**Problem Statement:**
A gas takes up a volume of 17 liters, has a pressure of 2.3 atm, and a temperature of 299 K. If I raise the temperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas?

**Solution Steps:**
1. Use the same combined gas law: \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \)
2. Substitute the values:
   - \( P_1 = 2.3 \, \text{atm} \)
   - \( V_1 = 17 \, \text{L} \)
   - \( T_1 = 299 \, \text{K} \)
   - \( P_2 = 1.5 \, \text{atm} \)
   - \( T_2 = 350 \
Transcribed Image Text:# Educational Text on Gas Laws ## Combined Gas Law Problems ### Example Problem 1: **Problem Statement:** If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas? **Solution Steps:** 1. Use the formula: \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \) 2. Substitute known values: - \( P_1 = 12 \, \text{atm} \) - \( V_1 = 23 \, \text{L} \) - \( T_1 = 200 \, \text{K} \) - \( P_2 = 14 \, \text{atm} \) - \( T_2 = 300 \, \text{K} \) 3. Solve for \( V_2 \): \[ V_2 = \frac{(12 \, \text{atm}) \times (23 \, \text{L}) \times (300 \, \text{K})}{(14 \, \text{atm}) \times (200 \, \text{K})} = 29.57 \, \text{L} \] ### Example Problem 2: **Problem Statement:** A gas takes up a volume of 17 liters, has a pressure of 2.3 atm, and a temperature of 299 K. If I raise the temperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas? **Solution Steps:** 1. Use the same combined gas law: \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \) 2. Substitute the values: - \( P_1 = 2.3 \, \text{atm} \) - \( V_1 = 17 \, \text{L} \) - \( T_1 = 299 \, \text{K} \) - \( P_2 = 1.5 \, \text{atm} \) - \( T_2 = 350 \
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