Question

A cylinder contains 0.0100 mol of helium at 27.0 ?C.

How much heat is needed to raise the temperature to 67.0 ?C while keeping the volume constant?(4.99)

If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0 ?C to 67.0 ?C?(8.31 J)

If the gas is ideal, what is the change in its internal energy in part A?

If the gas is ideal, what is the change in its internal energy in part B?

Answer 1

according to the first law of thermodynamics,

dq=du+dw

dq=du when volume is constant

dq=n*Cv*dT where n=no.of moles

Cv=molar specific heat capacity at constant volume

dT=change in temperature

dq=0.01*(3R/2)*40 R=gas constant

=4.99 J

when pressure is kept constant,

work done by the gas=nRdT Cp for monoatomic gas=5/2

=3.3256 J

dq=du+dw

dq=4.99+3.32

dq=8.31 J

change in internal energy is same in both cases and is=4.99 J