Internal energy change

Classical Mechanics Level 2

Monoatomic ideal gases with the same temperature and volume are filled in the same kind of cylinders [A] and [B], as shown above, and are being heated by the same quantity of heat $Q.$ The piston in [A] can move freely, whereas the piston in [B] is fixed. Let $\Delta U_A$ and $\Delta U_B$ denote the increases in the amount of internal energies of the gases in [A] and [B], respectively, then what is the ratio $\Delta U_A:\Delta U_B?$

Ignore the mass of the piston, and assume that the energy loss to the outside is negligible.

1:2

3:5

3:2

2:3

1 solution

Subham Jyoti Mishra
Apr 25, 2014

in cylinder A , U=nCvdeltaT=Q*Cv/Cp and in cylinder B, W=0; U=Q so taking the ratio; Ua/Ub=Cv/Cp=3/5 for a monoatomic gas.

For, cylinder A: del Q = delU+Work= n Cp deltaTa .... eq. (1)

whereas del Q=del U= n Cv delta Tb..... eq. (2) for cylinder B as the piston is fixed.

As the same amount of heat is supplied, the del Q is same in both cases, so delta Ta/ delta Tb= Cv/Cp if we divide the equation 1 by 2

implies delta Ta/ delta Tb= (3R/2)/(5R/2)=3/5.

Now delta U1/ delta U2 = n Cv delta Ta/ n Cv*delta Tb= delta Ta/ delta Tb= 3/5

Varun Goenka - 7 years, 1 month ago

Internal energy change

1 solution

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