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Four particles have speed 2, 3, 4 and 5 cm/s respectively. Their rms speed is:

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Explanation

4.

22 + 32 + 42 + 5244 + 9 + 16 + 25454/2

A gaseous mixture contains 4 molecules with a velocity of 6 cm sec-1, 5 molecules with a velocity of 2 cm sec-1 and 10 molecules with a velocity of 3 cm sec-1. What is the RMS velocity of the gas:

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Explanation

3.

4 × 62 + 5 + 22 + 10 + 3219     = 4 + 36 + 5 + 4 + 10 + 919     = 44 + 20 + 9019 = 25412     = 3.6 cm/sec-1

The ratio between the root mean square velocity of H2 at 50 K and that of O2 at 800 K is:

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If C1, C2, C3... represent the speeds of n1, n2, n3... molecules respectively, then the root mean square speed will be:

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Explanation

The root mean square velocity or vrmsvrms  is the square root of the average square velocity and is

 

 

The root mean square velocity of hydrogen is 5times than that of nitrogen. If T is the temperature of the gas, then:

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Explanation

Vrms=3RTM=VrmsH2VrmsN2=TH2MH2×MN2TN2;             VrmsH2=5VrmsN2           VrmsH2VrmsH2×5=TH2TN2×282                                          =51=TH2TN2×14                                          = 5=TH2TN2×14                            TN2×5=TH2×14                                TN2>TH2

At what temperature will average speed of the molecules of the second member of the series CnH2n be the same of Cl2 at 627°C?

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Explanation

Second memeber of CnH2n series    = C3H6=42    = 8RT1πM1=8RT2πM2=90071=T242             T2=532.4 K

If URMS of a gas is 30 R1/2 ms-1 at 27°C then the molar mass of gas is:

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Explanation

Urms=3RTMor  30R12=3RTMor  302=300×3MM=1.0 g mol-1=0.001 kg mol-1Hence answer is d

The compressibility factor for nitrogen at 330 K and 800 atm is 1.90 and at 570 K and 200 atm is 1.10. A certain mass of N2 occupies a volume of 1 dm3 at 330 K and 800 atm. Calculate volume occupied by same quantity of N2 gas at 570 K and 200 atm:

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Explanation

Z=PVnRT;    1.90=1×800n×R×330;                n=1×8001.90×R×330             Z=1.10=V×200n×R×570;                   1.10=V×200×1.90×R×330800×R×570                   V=4 L

Consider the following statements. If the van der Waals' parameters of two gases are given as

                                a/dm6 bar mol-2                               b/dm3 mol-1Gas A                                6.5                                                 0.055Gas B                                  2                                                     0.01

then:

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Pressure remaining the same, the volume of a given mass of an ideal gas increase for every degree centigrade rise in temperature by definite fraction of its volume at:

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Explanation

1.

The volume of a fixed mass of dry gas increases or decreases by ​1273 times the volume at 0 °C for every 1 °C rise or fall in temperature.

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