A gas is in a container with an initial volume of 2.50L, initial pressure of 1.34 atm, and initial temperature of 308 K. What would be the p

Question

A gas is in a container with an initial volume of 2.50L, initial pressure of 1.34 atm, and initial temperature of 308 K. What would be the pressure if the gas was heated to 373 K and the volume decreased to 1.90 L? *Hint use the Combined Gas Law.

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Tryphena 3 years 2021-08-25T05:04:49+00:00 1 Answers 5 views 0

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    2021-08-25T05:06:07+00:00

    Answer: The pressure is 2.14 atm if the gas was heated to 373 K and the volume decreased to 1.90 L.

    Explanation:

    Given: V_{1} = 2.50 L,      P_{1} = 1.34 atm,       T_{1} = 308 K

    P_{2} = ? ,            V_{2} = 1.90 L,          T_{1} = 373 K

    Formula used to calculate the final pressure is as follows.

    \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

    Substitute the values into above formula as follows.

    \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1.34 atm \times 2.50 L}{308 K} = \frac{P_{2} \times 1.90 L}{373 K}\\P_{2} = \frac{1.34 atm \times 2.50 L \times 373}{308 K \times 1.90 L}\\= 2.14 atm

    Thus, we can conclude that the pressure is 2.14 atm if the gas was heated to 373 K and the volume decreased to 1.90 L.

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