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1 mol of He at 57^@C is mixed with 1 mol...

1 mol of He at `57^@C` is mixed with 1 mol of Ar at `27^@C`. Find the temperature of the gas mixture .

Text Solution

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Average kinetic energy of 1 mol gas
`=3/2 RT =3/2kN_AT`
`therefore` Average kinetic energy of 1 mol of He
`=3/2kN_A(273+57)=3/2kN_A times 330`
Average kinetic energy of 1 mol of Ar
`=3/2 kN_A(273+27)=3/2KN_A times 330`
After mixing the two gases,the number of atoms in the mixture =`2N_A`
Let the temperatuer of the mixture is T.
Average kinetic energy of the mixture =`3/2 times 2N_AkT`
`therefore` From the principle of conservation of energy
`3/2times2N_AkT=3/2kN_Atimes330+3/2kN_Atimes300`
or,`T=1/2(330+300)=315K=(315-273)=42^@C`
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