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When 1 mole of gas is heated at constant...

When 1 mole of gas is heated at constant volume. Temperature is raised from `298` to `308K` . Heat supplied to the gas is `500J` . Then which stamenet is correct?

A

`q=-W=500J,DeltaU=0`

B

`q=DeltaU=500J,W=0`

C

`q=-W=500J,DeltaU=0`

D

`DeltaU=0,q=W=-500J`

Text Solution

Verified by Experts

The correct Answer is:
C
We Know that, `DeltaH=DeltaE+PDeltaV`
When, `DeltaV=0` , `:. DeltaH=DeltaE`
From the first law of thermodynamics,
`DeltaE=q-W`
In the given problem,`DeltaH=500J`
`-W=- PDeltaV,DeltaV=0`
So, `DeltaE=q=500J`
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Knowledge Check

  • When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct ?

    A
    `q = Deltacup = -500 J, w = 0`
    B
    `q = Deltacup = + 500 J, w = 0`
    C
    `q = w - 500 J, Deltacup = 0`
    D
    `Delta cup = 0, q = w = -500 J`

  • When 1 mole of a gas is heated at constant volume temperature is raised from 298K to 308 K heat supplied to the gas is 500J then which statement is correct?

    A
    `q=-W=500J, deltaE=0`
    B
    `q=W=500J, deltaE=0`
    C
    `q=deltaE=500J, W=0`
    D
    `deltaE=0,q=W=-500J`

  • When 1 mole of a gas is heated at constant volume temperature is raised from 298K to 308 K heat supplied to the gas is 500J then which statement is correct?

    A
    `q=-W=500J, deltaE=0`
    B
    `q=W=500J, deltaE=0`
    C
    `q=deltaE=500J, W=0`
    D
    `deltaE=0,q=W=-500J`

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