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3 years ago

K = °C + 273 A 4.1 L sample of gas is held at 25 °C. If the gas expands to 6.8 L, what is the final temperature?

Chemistry

1 answer:

gregori [183]3 years ago

6 0

Answer:

221 °C

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 4.1 L

Initial temperature (T₁) = 25 °C

= 25 °C + 273

= 298 K

Final volume (V₂) = 6.8 L

Final temperature (T₂) =?

The final temperature of the gas can be obtained as follow:

V₁ / T₁ = V₂ / T₂

4.1 / 298 = 6.8 / T₂

Cross multiply

4.1 × T₂ = 298 × 6.8

4.1 × T₂ = 2026.4

Divide both side by 4.1

T₂ = 2026.4 / 4.1

T₂ ≈ 494 K

Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:

°C = K – 273

K = 494

°C = 494 – 273

°C = 221 °C

Thus the final temperature of the gas is 221 °C

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<u>Explanation:</u>

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Given mass of carbon dioxide = 1 lb = 453.6 g (Conversion factor: 1 lb = 453.6 g)

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

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To calculate the temperature of gas, we use the equation given by ideal gas equation:

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n = number of moles of carbon dioxide = 10.31 mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the system = ?

Putting values in above equation, we get:

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