Question

5 l of an unkown ideal gas weigh 10g at 1.5 ATM pressure what will be the RMS speed of unknown gas molecule​

Answer 1

Answer:

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Answer 2

Answer:

477.33 m/s

Explanation:

We know PV = nRT for an ideal gas.

But 'n', the number of moles, can be expressed as W/M, where W is the weight of the gas provided and M is the molar mass of the gas.

Here W = 10g

Rewriting the ideal gas equation, PV = $\frac{W}{M} RT$

$\frac{T}{M} = \frac{PV}{WR}$

= (1.5 x 5)/(10 x 0.0821)

= 9.135 K/gm

= 9135 K/kg

The expression for RMS velocity is: V = $\sqrt{\frac{3RT}{M}}$

= $\sqrt{3 * 0.0821* 9135} = 477.33 m/s$

Therefore the RMS velocity of the unknown gas molecule is 477.33 m/s