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Maxwell–Boltzmann Distribution

The diagram s...

Question

The diagram shows the Maxwell’s speed distribution curves for a certain ideal gas at two different temperature $T_{1}$ and $T_{2}$ . Pick the correct options:
Take $R = \frac{1}{12} L atm/molK$

At 400 K, most probable speed is 800 m/s

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At 400 K, rms speed is 1000 m/s

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T_{2} = 600 K

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M = 5 g / m o l

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Solution

The correct options are
B At 400 K, rms speed is 1000 m/s
C $T_{2} = 600 K$
$c_{m p} = \sqrt{\frac{2 R T}{M}} c_{r m s} = \sqrt{\frac{3 R T}{M}}$
At $150 K, c_{m p} = 500 m s^{- 1} \Rightarrow 500 = \sqrt{\frac{2 R T}{M}} = \sqrt{\frac{2 \times 25 / 3 \times 150}{M}} \Rightarrow M = 10 \times 10^{- 3} k g = 10 g / m o l$
At $T_{2}, c_{m p} = 1000 m s^{- 1} \Rightarrow 1000 = \sqrt{\frac{2 \times 25 / 3 T_{2}}{10 \times 10 - 3}} \Rightarrow T_{2} = 600 K$
At $400 K, c_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 \times 25 / 3 \times 400}{10 \times 10^{- 3}}} = 1000 m s^{- 1}$

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Maxwell–Boltzmann Distribution

Standard XII Chemistry

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The diagram shows the Maxwell’s speed distribution curves for a certain ideal gas at two different temperature T1 and T2. Pick the correct options: Take R=112 L atm/molK

The correct options are B At 400 K, rms speed is 1000 m/s C T2=600 Kcmp=√2RTMcrms=√3RTM At 150 K, cmp=500 ms−1⇒500=√2RTM=√2×25/3×150M⇒M=10×10−3 kg=10 g/mol At T2,cmp=1000 m s−1⇒1000=√2×25/3T210×10−3⇒T2=600 K At 400 K,crms=√3RTM=√3×25/3×40010×10−3=1000 m s−1

The diagram shows the Maxwell’s speed distribution curves for a certain ideal gas at two different temperature $T_{1}$ and $T_{2}$ . Pick the correct options:
Take $R = \frac{1}{12} L atm/molK$