Ask questions and get helpful answers.

A spherical glass container of unknown volume contains helium gas at 25C and 1.960 atm. When a portion of the helium is withdrawn and adjusted to 1.00 atm at 25C, it is found to have a volume of 1.75 cm^3. The fas remaining in the first container shows a pressure of 1.710 atm. Calculate the volume of the spherical container.

Use PV = nRT to calculate n for the gas removed.
Then substitute this n in PV = nRT for the shperical container using P = 1.960 - 1.710 and solve for V.
Post your work if you get stuck.

1 atm X 1.75 cm^3 = n X(.08206)X 298.15K

n = 0.0715?
For the container.

(1.960 - 1.710 atm) X Volume = 0.0715 X (0.08206) X 298.15K.

Volume = 0.1429 cm^3?

That small?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩

2 answers

  1. the stepsare correct but the arthimetic is wrong somewhere in between. You should get 7 ml

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. agree with dmoney, the n value(moles) does not equal 0.0715, its 7.156 ^-3 which would make the answer 6.9938... or 7mL. Do not convert from mL to L in between, either do it at the beginning or the end.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.