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A gas is at 35.0°C and 3.00 L. What is the temperature of the gas if the volume is increased to 6 00 L?

O 616°C
O 17.5°C
O 1.16 °C
O 65.0°C
O 343°C

1 Answer

2 votes

Final answer:

The new temperature of the gas after the volume is doubled from 3.00 L to 6.00 L, according to Charles's law, is 343°C. option e is correct

Step-by-step explanation:

option e is correct The student's question involves calculating the new temperature of a gas when its volume is doubled, assuming constant pressure and amount of gas, which is an application of Charles's law. Charles's law states that the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature in Kelvins. To find the new temperature when the gas volume increases from 3.00 L to 6.00 L, we start by converting the initial temperature from 35.0°C to Kelvins (35.0 + 273.15 = 308.15 K), then we use the proportionality to solve for the final temperature (T2).

V1/T1 = V2/T2 → T2 = (V2 × T1) / V1 → T2 = (6.00 L × 308.15 K) / 3.00 L → T2 = 616.30 K. Converting back to Celsius gives us T2 = 616.30 K - 273.15 = 343.15°C, which can be rounded to 343°C.

According to Charles's Law, the volume of a gas is directly proportional to its temperature.

To find the final temperature of the gas when the volume is increased to 600 L, we can set up a proportion using the initial volume and temperature and the final volume:

(Initial Temperature) / (Final Temperature) = (Initial Volume) / (Final Volume)

Plugging in the values given, we have:

35.0°C / (Final Temperature) = 3.00 L / 600 L

Simplifying the equation, we get: Final Temperature = (35.0°C * 600 L) / 3.00 L = 7000 °C

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