Question

The temperature at which nitrogen under one atmospheric pressure has the same RMS velocity of carbon dioxide at STP is

Answer 1

Answer : 4570C

Explanation :

Vm=3RTM−−−−√

VN2VO2=1

3RT1MN2−−−−−√=3RT2MO2−−−−−√

T1=T2×MN2MO2

=400×2832=350K

Therefore , t=350−273= 770 C

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Answer 2

The temperature at which nitrogen under one atmospheric pressure has the same RMS velocity as carbon dioxide at STP is -99.27°C.

Given,

Two gases are Nitrogen and Carbon dioxide.

To Find,

The temperature at which nitrogen under one atmospheric pressure has the same RMS velocity as carbon dioxide at STP.

Solution,

The method of finding the required temperature is as follows -

According to the question, the RMS velocity of Nitrogen and Carbon dioxide is the same at STP at the required temperature.

So, $U_{N_{2}}=U_{CO_{2}}$ , where $U_{N_{2}}$ and $U_{CO_{2}}$ are the RMS velocity of Nitrogen and Carbon dioxide respectively.

Now, $\sqrt{ \frac{3RT_{1}}{M_{1}} }= \sqrt{\frac{3RT_{2}}{M_{2}}}$ [ where $T_{1}$ and $T_{2}$ are the temperature of  Nitrogen and Carbon dioxide respectively, R is the universal gas constant, and $M_{1}$  and $M_{2}$ are molar masses of Nitrogen and Carbon dioxide respectively ]

⇒ $\frac{3RT_{1}}{M_{1}} }= \frac{3RT_{2}}{M_{2}}}$ [Squaring on both sides]

⇒  $\frac{T_{1}}{M_{1}} }= \frac{T_{2}}{M_{2}}}$

⇒  $\frac{T_{1}}{28 }= \frac{273}{44}$ [Since at STP $T_{2}=273K$, the value of $M_{1}$ and $M_{2}$ are 28 and 44 respectively]

⇒  $T_{1}=\frac{273*28}{44} = 173.73$ (approximately)

So, the temperature is 173.73K = (173.73 -273) °C = -99.27 °C.

Hence, the temperature at which nitrogen under one atmospheric pressure has the same RMS velocity as carbon dioxide at STP is -99.27°C.

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