The Kelvin temperature required for 0.0470 mol of helium gas to fill a balloon to 1.20 L under 0.878 atm is 14.4 K

Question

The Kelvin temperature required for 0.0470 mol of helium gas to fill a balloon to 1.20 L under 0.878 atm is

14.4 K

273 K

298 K

307 K

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bonexptip 5 years 2021-08-25T02:49:19+00:00 1 Answers 3 views 0

Answers ( )

  1. thongdat2
    0
    2021-08-25T02:50:44+00:00
    Reply

    From the ideal gas law, PV = nRT, we can rearrange the equation to solve for T given the other parameters.

    T = PV/nR

    where P = 0.878 atm, V = 1.20 L, n = 0.0470 moles, and R = 0.082057 L•atm/mol•K. Plugging in our values, we obtain the temperature in Kelvin:

    T = (0.878 atm)(1.20 L)/(0.0470 mol)(0.082057 L•atm/mol•K)
    T = 273 K

    So, the second answer choice would be correct.

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