Ask Question
18 January, 21:45

A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What is the final pressure in atm?

+2
Answers (1)
  1. 18 January, 22:23
    0
    0.681 atm

    Explanation:

    To solve this problem, we make use of the General gas equation.

    Given:

    P1 = 785 torr

    V1 = 2L

    T1 = 37 = 37 + 273.15 = 310.15K

    P2 = ?

    V2 = 3.24L

    T2 = 58 = 58+273.15 = 331.15K

    P1V1/T1 = P2V2/T2

    Now, making P2 the subject of the formula,

    P2 = P1V1T2/T1V2

    P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

    P2 = 515.715 Torr

    We convert this to atm: 1 torr = 0.00132 atm

    515.715 Torr = 515.715 * 0.00132 = 0.681 atm
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What is the final ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers