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The ratio of translational and rotationa...

The ratio of translational and rotational kinetic energies at `100K` temperature is `3:2`. Then the internal energy of one mole gas at that temperature is `(R=8.3j//"mol"-K]`

A

`1175J`

B

`1037.5J`

C

`2075J`

D

`4150J`

Text Solution

AI Generated Solution

To find the internal energy of one mole of gas at a temperature of 100 K, given that the ratio of translational to rotational kinetic energies is 3:2, we can follow these steps: ### Step 1: Understand the relationship between kinetic energies and degrees of freedom The translational kinetic energy (TKE) and rotational kinetic energy (RKE) can be expressed in terms of their respective degrees of freedom. For a monatomic ideal gas, the translational degrees of freedom (F_t) is 3, and for a diatomic or polyatomic gas, the rotational degrees of freedom (F_r) can vary. ### Step 2: Set up the ratio of kinetic energies Given the ratio of translational to rotational kinetic energies is 3:2, we can express this as: \[ ...
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Knowledge Check

  • The internal energy of one mole of the diatomic gas at 200 K is

    A
    100 R
    B
    200 R
    C
    500 R
    D
    400 R

  • At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this, at higher temperature

    A
    `C_(V) = (3 R)/(2)` for a monatomic gas
    B
    `C_(V) gt (3 R)/(2)` for a monatomic gas
    C
    `C_(V) lt (3 R)/(2)` for a diatomic gas
    D
    `C_(V) gt (3 R)/(2)` for a diatomic gas

  • Mean kinetic energy of a perfect gas per gram at the temperature T K is

    A
    1/2 KT
    B
    KT
    C
    `(3)/(2) (RT)/(M)`
    D
    `(1)/(2)(RT)/(M)`

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