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The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200 K is \mathrm{v}_0 

. Then, r.m.s. speed of the gas at 800 K will be:

Option: 1

4 \mathrm{v}_0


Option: 2

2 \mathrm{v}_0


Option: 3

\mathrm{v}_0


Option: 4

\frac{v_0}{4}


Answers (1)

best_answer

using vrms       =\sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}}}
 

  \mathrm{v}_0 =\sqrt{\frac{3 \mathrm{R} \times 200}{\mathrm{~m}}}        .....(1)

(v’)  =   \sqrt{\frac{3 \mathrm{R} \times 800}{\mathrm{~m}}}       .......(2)

dividing (2) by (1)

\frac{\mathrm{v}^{\prime}}{\mathrm{v}_0}=\sqrt{\frac{800}{200}}=\sqrt{4}=2

or v’ = 2v0

Posted by

rishi.raj

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