JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 14)
A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27oC. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/K mole]
0.9 kJ
6 kJ
10 kJ
14 kJ
Explanation
We know,
Vrms $$ \propto $$ $$\sqrt T $$
So, to make Vrms double we have to make temperature 4 times.
$$ \therefore $$ Final temperature = 300 $$ \times $$ 4 = 1200 K
As N2 gas present in the closed vessel
So it is a isochoric process.
$$ \therefore $$ Q = nCv $$\Delta $$ T
= $${{15} \over {28}} \times \left( {{5 \over 2}R} \right)\left( {1200 - 300} \right)$$
= 10000 J
= 10 kJ
Vrms $$ \propto $$ $$\sqrt T $$
So, to make Vrms double we have to make temperature 4 times.
$$ \therefore $$ Final temperature = 300 $$ \times $$ 4 = 1200 K
As N2 gas present in the closed vessel
So it is a isochoric process.
$$ \therefore $$ Q = nCv $$\Delta $$ T
= $${{15} \over {28}} \times \left( {{5 \over 2}R} \right)\left( {1200 - 300} \right)$$
= 10000 J
= 10 kJ
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