Question

at what temperature will root mean square speed of molecule of an ideal gas be thrice the root mean square speed at temperature 40 Kelvin ​

Answer 1

Answer:

360K

C=(3RT/M)^1/2

Explanation:

C1/C2=(T1/T2)^1/2

C/3C=(40/T2)^1/2

1/9=40/T2

T2=360K

Answer 2

Answer:

Root mean square speed of Argan atom

( Vrms1 ) = √{3RT1/M1} -----(1)

Root mean square spped of Helium atom ( Vrms2) = √ { 3RT2/M2}-------(2)

Divide equations (1) and (2)

Vrms1/Vrms2 = √{3RT1/M1}/√{3RT2/M2}

= √{ T1 ×M2/T2×M2}

But ,

Vrms1 = Vrms2 [ A/C to question , ]

1 = √{ T1 × M2/M1 × T2}

M1/M2 = T1/T2

So,

T1 = T2 × { M1/M2}

Here,

T2 = -20°C = -20+273 = 253 K

M1 = 40g/mol

M2 = 4 g/mol

T1 = ?

T1 = 253 × { 40 /4 }

= 2530 K

Explanation: