The average speed at T1 K and most probable speed at T2 K of CO2 gas is 9 - 104 cm-sec- The values of T1 and T2 are

The average speed at $T_{1} K$ and most probable speed at $T_{2} K$ of $C O_{2}$ gas is $9 \times 10^{4}$ cm/sec. The values of $T_{1}$ and $T_{2}$ are

Q. The average velocity at

T_{1} K

, and the most probable velocity at

T_{2} K

C O_{2}

gas is

9.0 \times 10^{4} {cm sec}^{- 1}

. The value of

T_{1}

is K and

T_{2}

is K.

Q. For

C O_{2}

, given that average velocity at

T_{1}

is equal to most probable velocity at

T_{2}

. Thus,

\frac{T_{1}}{T_{2}}

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The average speed at T1 K and most probable speed at T2 K of CO2 gas is 9×104 cm/sec. The values of T1 and T2 are

The correct option is C 1684K; 2143K ¯C=Cp √8RT1Mπ=√2RTM 8T1π=T2 T1T2=π8=22718=2256 T1=1684K T2=2143K

The average speed at $T_{1} K$ and most probable speed at $T_{2} K$ of $C O_{2}$ gas is $9 \times 10^{4}$ cm/sec. The values of $T_{1}$ and $T_{2}$ are

The correct option is C
1684K; 2143K

$¯ C = C_{p}$

$\sqrt{\frac{8 R T_{1}}{M π}} = \sqrt{\frac{2 R T}{M}}$

$\frac{8 T_{1}}{π} = T_{2}$

$\frac{T_{1}}{T_{2}} = \frac{π}{8} = \frac{22}{7} \frac{1}{8} = \frac{22}{56}$

$T_{1} = 1684 K$

$T_{2} = 2143 K$