JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 20)
An ideal gas occupies a volume of 2m3 at a pressure of 3 $$ \times $$ 106 Pa. The energy of the gas is :
6 $$ \times $$ 104 J
9$$ \times $$ 106 J
3 $$ \times $$ 102 J
108 J
Explanation
Energy = $${1 \over 2}$$ nRT = $${f \over 2}$$PV
= $${f \over 2}$$ (3 $$ \times $$ 106) (2)
= f $$ \times $$ 3 $$ \times $$ 106
Considering gas is monoatomic i.e. f = 3
E. = 9 $$ \times $$ 106 J
= $${f \over 2}$$ (3 $$ \times $$ 106) (2)
= f $$ \times $$ 3 $$ \times $$ 106
Considering gas is monoatomic i.e. f = 3
E. = 9 $$ \times $$ 106 J
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