JEE MAIN - Physics (2006 - No. 25)
The work of $$146$$ $$kJ$$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by $${7^ \circ }C.$$ The gas is $$\left( {R = 8.3J\,\,mo{l^{ - 1}}\,{K^{ - 1}}} \right)$$
diatomic
triatomic
a mixture of monoatomic and diatomic
monoatomic
Explanation
$$W = {{nR\Delta T} \over {1 - \gamma }} \Rightarrow - 146000$$
$$ = {{1000 \times 8.3 \times 7} \over {1 - \gamma }}$$
or $$1 - \gamma = - {{58.1} \over {146}} \Rightarrow \gamma $$
$$ = 1 + {{58.1} \over {146}} = 1.4$$
Hence the gas is diatomic.
$$ = {{1000 \times 8.3 \times 7} \over {1 - \gamma }}$$
or $$1 - \gamma = - {{58.1} \over {146}} \Rightarrow \gamma $$
$$ = 1 + {{58.1} \over {146}} = 1.4$$
Hence the gas is diatomic.
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