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0.014 Kg of N2 gas at 27 0C is kept in a closed vessel. How much heat is required to double the rms speed of the N2 molecules?

Option: 1

3000 cal


Option: 2

2250 cal


Option: 3

2500 cal


Option: 4

3500 cal


Answers (1)

best_answer

Heat required, 

\Delta q=n C_v \Delta T

Here, 

n=\frac{0.014 \times 1000}{28}=\frac{1}{2}

For diatomic molecule, C_{v}=\frac{5}{2}R

and rms speed of N2 molecule, C\alpha \sqrt{T}

Therefore, to double the velocity c, temperature should be 4T.

\begin{aligned} \Delta Q= & \frac{1}{2} \times \frac{5}{2} R \times(47-T) \\ & =\frac{5}{4} \times 2 \times(273+27) \times 3=2250 \mathrm{cal} \end{aligned}

Posted by

Rakesh

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