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52.1 L of a gas is stored at 23.0°C and 700.0 mmHg. If the temperature is increased to 401 K and the pressure is dropped to 400.0 mmHg, what is the new volume of the gas?

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Final answer:

Using the combined gas law, the new volume of the gas when the temperature is raised to 401 K and the pressure is reduced to 400.0 mmHg is calculated to be 30.76 L.

Step-by-step explanation:

To calculate the new volume of the gas when the temperature and pressure are changed, we can use the combined gas law, which is derived from Boyle's Law, Charles's Law, and Gay-Lussac's Law. The combined gas law formula is:

V1/T1 * P1 = V2/T2 * P2

Where:

  • V1 is the initial volume of the gas
  • T1 is the initial temperature in Kelvin
  • P1 is the initial pressure
  • V2 is the final volume of the gas
  • T2 is the final temperature in Kelvin
  • P2 is the final pressure

Firstly, we need to convert all the values to appropriate units:

  • Initial temperature (T1) needs to be converted from °C to K: 23.0 °C = 296.15 K
  • Initial pressure (P1) is given in mmHg, which is fine for these calculations
  • Final temperature (T2) is already given in Kelvin
  • Final pressure (P2) is also in mmHg, which is suitable

Applying the combined gas law and solving for V2, the new volume:

(52.1 L / 296.15 K) * 700.0 mmHg = V2 / 401 K * 400.0 mmHg

V2 = (52.1 L * 700.0 mmHg * 401 K) / (296.15 K * 400.0 mmHg)

Next, calculate the value:

V2 = (3644700 L*mmHg*K) / (118460 K*mmHg)

V2 = 30.76 L

Therefore, the new volume of the gas when the temperature is increased to 401 K and the pressure is decreased to 400.0 mmHg is 30.76 L.

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