Q. 14.5( 2 Votes )

The pressure p and volume V of an ideal gas both increase in a process.
A. Such a process is not possible.

B. The work done by the system is positive.

C. The temperature of the system must increase.

D. Heat supplied to the gas is equal to the change in internal energy.

Answer :

We know that,


Work done = force ×displacement



Volume = area ×displacement


Therefore,


Work done=pressure ×volume


Let change in the volume of system = ΔV = V2-V1


Pressure =P


Thus, work done by the system W


ΔW=PΔV


It is given that volume in increasing i.e. ΔV>0. So, ΔW>0 and positive.


From ideal gas equation, we know that


PV=nRT


Where P=pressure


V=volume


n=number of moles


R=gas constant


T=temperature


So, if both pressure and volume are increasing, then


the temperature must also increase as n and R are constant.

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