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When 3.0 mole of an ideal diatomic gas i...

When 3.0 mole of an ideal diatomic gas is heated and compressed simultaneously from 300K, 1.0 atm to 400K and 5.0atm, the change in entropy is (Use `C_(P) = (7)/(2)R` for the gas)

A

`-20 JK^(-)`

B

`-5 JK^(-)`

C

`-15 JK^(-)`

D

`-2.8 JK^(-)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta S = nC_(P) "ln" (T_(2))/(T_(1)) nR "ln" (P_(2))/(P_(1))`
`= 3 xx (7)/(2) R xx "ln" (5)/(1) = -15`
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Knowledge Check

  • Calculate Delta S for 3 moles of a diatomic ideal gas which is heated and compressed from 298K and 1bar to 596K and 4 bar

    A
    `-14.7 "cal" K^(-1)`
    B
    `+14.7 "cal" K^(-1)`
    C
    `-4.9 "cal" K^(-1)`
    D
    `6.3 "cal" K^(-1)`

  • When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas (Delta S) is

    A
    `C_(p, m) ln 2`
    B
    `C_(v, m) ln 2`
    C
    R ln 2
    D
    `(C_(v, m)- R) ln 2`

  • The weight of 350 ml of a diatomic gas is 1 gram at 0^@C and 2 atm. pressure. The weight of one atom in grams is

    A
    `32/N`
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    `16/N xx 1/2`
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    `2.66 xx 10^(-23)`
    D
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