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The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3ml^-1Jmol^-1K^-1)`A. diatomicB. triatomicC. A mixture of monoatomic and diatomicD. monoatomic |
Answer» Correct Answer - A Work done in adiabatic process `W = (muR(T_(1)-T_(2)))/(gamma-1) implies gamma = 1 + (R(T_(2) -T_(1)))/(W)` =`1 + (10^(3) xx 8.3(7))/(146 xx 10^(3)) = 1 + 0.40 = 1.40` `therefore ` The gas must be diatomic |
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