1.

In the process `pV^2=` constant, if temperature of gas is increased, thenA. (a) change in internal energy of gas is positiveB. (b) work done by gas is positiveC. (c) heat is given to the gasD. (d) heat is taken out from the gas

Answer» Correct Answer - A::C
Temperature is increased. So, internal energy will also increase.
`:.` `DeltaU=+ve`
Further,
`pV^2=` constant
`:.` `((T)/(V))V^2=`constant or `Vprop1/T`
Temperature is increased. So, volume will decrease and work done will be negative.
In the process `pV^x`= constant, molar heat capacity is given by
`C=C_V+(R)/(1-x)`
Here, `x=2`
`:.` `C=C_V-R`
`C_V` of any gas is greater than R.
So, C is positive. Hence, form the equation,
`Q=nCDeltaT`
Q is positive if T is increased.
or `DeltaT` is positive.


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