A comple of cartna dioxide occupies a volume of 3.50 L at 125 kPa pressure. What pressure moveld the gas exert if the volume was decreased to 2.0 L ?

Question

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We are given:
  
- $$ P_1 = 125 \, \text{kPa} $$  
- $$ V_1 = 3.50 \, \text{L} $$  
- $$ V_2 = 2.0 \, \text{L} $$

Since the gas is compressed at constant temperature, we apply Boyle's law:

$$
P_1 V_1 = P_2 V_2
$$

Solving for $$ P_2 $$:

$$
P_2 = \frac{P_1 V_1}{V_2}
$$

Substitute the given values:

$$
P_2 = \frac{125 \, \text{kPa} \times 3.50 \, \text{L}}{2.0 \, \text{L}}
$$

Calculate the numerator:

$$
125 \, \text{kPa} \times 3.50 \, \text{L} = 437.5 \, \text{kPa} \cdot \text{L}
$$

Now divide by $$ 2.0 \, \text{L} $$:

$$
P_2 = \frac{437.5 \, \text{kPa} \cdot \text{L}}{2.0 \, \text{L}} = 218.75 \, \text{kPa}
$$

Thus, the pressure exerted by the gas at a volume of $$ 2.0 \, \text{L} $$ is approximately:

$$
218.75 \, \text{kPa}
$$
The pressure increases to 218.75 kPa when the volume is decreased to 2.0 L.

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