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Standard XII

Chemistry

First Law of Thermodynamics

A cylinder pi...

Question

A cylinder piston arrangement has n molecules at an absolute temperature T. If the number of molecules in the arrangement is doubled, the total kinetic energy of all the molecules remains the same as before. The ratio of the temperature of the arrangement before and now.

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Solution

Total energy of n molecule= $\frac{1}{2} n K T_{1}$
Now, if $n' = 2 n$
then $E' = \frac{1}{2} \times 2 n \times 4 \times K T$
$\frac{1}{2} n K T_{1}$ $= \frac{1}{2} \times 2 n \times K \times T_{2}$
$\frac{T_{1}}{T_{2}} = 2$
$∴ T_{1} : T_{2} = 2 : 1$

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A cylinder piston arrangement has n molecules at an absolute temperature T. If the number of molecules in the arrangement is doubled, the total kinetic energy of all the molecules remains the same as before. The ratio of the temperature of the arrangement before and now.

Total energy of n molecule=12nKT1Now, if n′=2nthen E′=12×2n×4×KT12nKT1=12×2n×K×T2T1T2=2∴T1:T2=2:1

Total energy of n molecule= $\frac{1}{2} n K T_{1}$
Now, if $n' = 2 n$
then $E' = \frac{1}{2} \times 2 n \times 4 \times K T$
$\frac{1}{2} n K T_{1}$ $= \frac{1}{2} \times 2 n \times K \times T_{2}$
$\frac{T_{1}}{T_{2}} = 2$
$∴ T_{1} : T_{2} = 2 : 1$