Find the most probable speed V mp of nitrogen gas molecules at T -560 K- Given- - R -2-88- where R is gas constantA- 460 m - sB- 500 m - sC- 576 m - sD- 600 m - s

The correct option is C 576 m/sWe know that, V_mp=√(2RTM) ⇒ V_mp=√(2× R× 56028× 10^ 3) nbsp; (∵ Molar mass of N_2=28× 10^ 3 kg ) ⇒ V_mp=200√(R)=576 m/s ...

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Standard XII

Physics

Root Mean Square Speed

Find the most...

Question

Find the most probable speed $(V_{m p})$ of nitrogen gas molecules at $T = 560 K$ . (Given: $\sqrt{R} = 2.88$ , where $R$ is gas constant)

460 m/s

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500 m/s

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576 m/s

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600 m/s

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Solution

The correct option is C $576 m/s$
We know that,
$V_{m p} = \sqrt{\frac{2 R T}{M}}$
$\Rightarrow V_{m p} = \sqrt{\frac{2 \times R \times 560}{28 \times 10^{- 3}}}$
( $∵$ Molar mass of $N_{2} = 28 \times 10^{- 3} kg$ )
$\Rightarrow V_{m p} = 200 \sqrt{R} = 576 m/s$

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Root Mean Square Speed

Standard XII Physics

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Find the most probable speed (Vmp) of nitrogen gas molecules at T=560 K. (Given: √R=2.88, where R is gas constant)

The correct option is C 576 m/sWe know that, Vmp=√2RTM ⇒Vmp=√2×R×56028×10−3 (∵ Molar mass of N2=28×10−3 kg ) ⇒Vmp=200√R=576 m/s

Find the most probable speed $(V_{m p})$ of nitrogen gas molecules at $T = 560 K$ . (Given: $\sqrt{R} = 2.88$ , where $R$ is gas constant)

The correct option is C $576 m/s$
We know that,
$V_{m p} = \sqrt{\frac{2 R T}{M}}$
$\Rightarrow V_{m p} = \sqrt{\frac{2 \times R \times 560}{28 \times 10^{- 3}}}$
( $∵$ Molar mass of $N_{2} = 28 \times 10^{- 3} kg$ )
$\Rightarrow V_{m p} = 200 \sqrt{R} = 576 m/s$