10 moles of an ideal monoatomic gas at `10^(@)C` are mixed with 20 moles of another monoatomic gas at `20^(@)C`. Then the temp. of the mixture is

Options

A. `15.5^(@)C`
B. `15^(@)C`
C. `16^(@)C`
D. `16.6^(@)C`

Answer

D

Explanation

As both the gases ae monoatomic, their molar specific heat is same, say C.
Let T be the temp.of mixture.
As heat gained by one = heat lost by other
`:. 10 xx C (T-10) = 20 xx C(20-T)`
`T-10 = 40-2T`
`3T = 50, T = 16.6^(@)C`.