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In the process pV^2= constant, if temper...

In the process `pV^2=` constant, if temperature of gas is increased, then

A

(a) change in internal energy of gas is positive

B

(b) work done by gas is positive

C

(c) heat is given to the gas

D

(d) heat is taken out from the gas

Text Solution

Verified by Experts

The correct Answer is:
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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