Question

The temperature of a flexible container of gas si raised from 100.0 C° to 175 C° and its pressure changes from 650.0 torr to 925 torr causing a new volume of 235 mL. What was the original volume of the container of gas?

Answer 1

To solve this problem, you can use the combined gas law, which is expressed as:

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

where:

-
`\( P_1 \) and \( P_2 \)` are the initial and final pressures,

-
`\( V_1 \) and \( V_2 \)` are the initial and final volumes,

-
`\( T_1 \) and \( T_2 \)` are the initial and final temperatures.

First, make sure temperatures are in Kelvin, as the temperature must be in Kelvin for gas law calculations.

`\[ T_1 = 100.0 + 273.15 = 373.15 \, \text{K} \]`

`\[ T_2 = 175 + 273.15 = 448.15 \, \text{K} \]`

Now, plug in the values:

`\[ \frac{650.0 \, \text{torr} \cdot V_1}{373.15 \, \text{K}} = \frac{925 \, \text{torr} \cdot 235 \, \text{mL}}{448.15 \, \text{K}} \]`

Now, solve for
`\( V_1 \)`, the original volume.

`\[ 650.0 \, \text{torr} \cdot V_1 = \frac{925 \, \text{torr} \cdot 235 \, \text{mL} \cdot 373.15 \, \text{K}}{448.15 \, \text{K}} \]`

`\[ V_1 = \frac{925 \, \text{torr} \cdot 235 \, \text{mL} \cdot 373.15 \, \text{K}}{650.0 \, \text{torr} \cdot 448.15 \, \text{K}} \]`

`\[ V_1 \approx 279.6 \, \text{mL} \]`

So, the original volume of the container of gas was approximately
`\(279.6 \, \text{mL}\)`.

Answer 2

(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂

Where:
P₁ and P₂ are the initial and final pressures,
V₁ and V₂ are the initial and final volumes,
T₁ and T₂ are the initial and final temperatures.

Given:
P₁ = 650.0 torr
P₂ = 925 torr
V₂ = 235 mL
T₁ = 100.0 °C + 273.15 (converting to Kelvin)
T₂ = 175 °C + 273.15 (converting to Kelvin)

We can plug these values into the combined gas law equation and solve for V₁, the original volume.

(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂

(650.0 torr × V₁) / (100.0 °C + 273.15) = (925 torr × 235 mL) / (175 °C + 273.15)

Now, let's calculate it:

(650.0 torr × V₁) / 373.15 K = (925 torr × 235 mL) / 448.15 K

Cross-multiplying:

650.0 torr × V₁ × 448.15 K = 925 torr × 235 mL × 373.15 K

650.0 torr × V₁ × 448.15 = 925 torr × 235 mL × 373.15

Now, we can solve for V₁:

V₁ = (925 torr × 235 mL × 373.15) / (650.0 torr × 448.15)

V₁ ≈ 198.8 mL

Therefore, the original volume of the container of gas was approximately 198.8 mL.