If at NTP velocity of sound in a gas is 1150 m / s, then the rms velocity of gas molecules at NTP is (Given: R=8.3 J / mol K, Cp=4.8 cal / mol K )

Question

If at NTP velocity of sound in a gas is 1150 m / s, then the rms velocity of gas molecules at NTP is (Given: R=8.3 J / mol K, Cp=4.8 cal / mol K ) (A) 1600 m/s (B) 1532.19 m/s (C) 160 m/s (D) zero

Tardigrade Admin · Accepted Answer

R=(8.3/4.2) cal g / mol K CV =Cp-R =(4.8-(8.3/4.2))=2.824 &gamma; =(Cp/CV)=(4.8/2.824)=1.69 Since, v=√((3/&gamma;)) × vs =√(3/1.69) × 1150 =1532.19 m / s

Q. If at NTP velocity of sound in a gas is $1150 m / s$ , then the rms velocity of gas molecules at NTP is (Given : $R = 8.3 J / m o l K, C_{p} = 4.8 c a l / m o l K$ )

1600 m/s

1532.19 m/s

160 m/s

zero

$R = \frac{8.3}{4.2} c a l g / m o l K$
$C_{V} = C_{p} - R$
$= (4.8 - \frac{8.3}{4.2}) = 2.824$
$γ = \frac{C _{p}}{C _{V}} = \frac{4.8}{2.824} = 1.69$
Since, $v = (\frac{3}{γ}) \times v_{s}$
$= \frac{3}{1.69} \times 1150$
$= 1532.19 m / s$

Q. If at NTP velocity of sound in a gas is 1150m/s, then the rms velocity of gas molecules at NTP is (Given : R=8.3J/molK,Cp​=4.8cal/molK )