Question

A vessel holds 250 mL of argon gas. If the pressure changes to 12 ATM and the temperature remains at 20 C, what is the new pressure of the gas?

Answer 1

To find the new pressure using the combined gas law, which states
`(P_1V_1)/(T_1) = (P_2V_2)/(T_2)`, where P is pressure, V is volume, and T is temperature. The new pressure of the argon gas is approximately 117.26 ATM.

Step-by-step explanation:

To find the new pressure using the combined gas law, which states
`(P_1V_1)/(T_1) = (P_2V_2)/(T_2)`, where P is pressure, V is volume, and T is temperature, you can rearrange the formula to solve for P₂:

`\[ P_2 = (P_1 \cdot V_1 \cdot T_2)/(V_2 \cdot T_1) \]`

Given that:

- P₁ (initial pressure) = 1 ATM,

- V₁ (initial volume) = 250 mL,

- T₁ (initial temperature) = 20°C + 273.15 (convert to Kelvin) = 293.15 K,

- T₂ (final temperature) = 20°C + 273.15 = 293.15 K,

- V) (final volume) remains the same, so V₂ = V₁.

Substitute these values into the formula:

`\[ P_2 = \frac{1 \, \text{ATM} \cdot 250 \, \text{mL} \cdot 293.15 \, \text{K}}{250 \, \text{mL} \cdot 293.15 \, \text{K}} \]`

`\[ P_2 = (29315)/(250) \, \text{ATM} \]`

P₂ = 117.26 ATM

Therefore, the new pressure of the argon gas is approximately 117.26 ATM.