Answer:

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas at different conditions.

The combined gas law is given by:

(P1 × V1) ÷ (T1) = (P2 × V2) ÷ (T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

We can rearrange this equation to solve for V2:

V2 = (P1 × V1 × T2) ÷ (P2 × T1)

Substituting the given values:

P1 = 0.80 atm

V1 = 60.0 mL

T1 = 60°C + 273.15 = 333.15 K

P2 = 0.50 atm

T2 = 50°C + 273.15 = 323.15 K

V2 = (0.80 atm × 60.0 mL × 323.15 K) ÷ (0.50 atm × 333.15 K)

V2 = 58.3 mL

Therefore, the final volume of the gas is 58.3 mL (rounded to three significant figures).

Verified answer

Answer:

Using the combined gas law, we have:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Where:

P₁ = 0.80 atm (initial pressure)

V₁ = 60.0 mL (initial volume)

T₁ = 60°C + 273.15 = 333.15 K (initial temperature in Kelvin)

P₂ = 0.50 atm (final pressure)

T₂ = 50°C + 273.15 = 323.15 K (final temperature in Kelvin)

V₂ = ? (final volume)

Plugging in the values, we get:

(0.80 atm * 60.0 mL) / 333.15 K = (0.50 atm * V₂) / 323.15 K

Simplifying the equation and solving for V₂, we get:

V₂ = (0.80 atm * 60.0 mL * 323.15 K) / (0.50 atm * 333.15 K) = 74.9 mL

Therefore, the final volume of the gas is 74.9 mL.

Explanation:

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