Question

47. Calculate Kinetic energy in Joules of 5 moles of nitrogen at 27°C.
49. Find the RMS velocity of CO2 gas at 27°C.​

Answer 1

Need to FinD :-

1. The Kinetic Energy.
2. The RMS Velocity.

• Question 47 :

$\red{\frak{Given}}\begin{cases}\sf The \ no.\ of \ moles \ = \ 5 \\\sf Temperature = 27^{\circ} \end{cases}$

The Kinetic Energy can be obtained by the following formula :-

$\dashrightarrow\red{\sf KE = \dfrac{3nRT}{2}}$

• Here n is the no. of moles ,T is the temperature in Kelvin and R is the constant whose value is 8.314 J / mol.K . Substituting the respective values ,

$\dashrightarrow\sf KE = \dfrac{3\times 8.314 \times 5 \times (27 + 273) }{2}\\\\\sf\dashrightarrow KE = \dfrac{3\times 8.314 \times 5 \times 300}{2}\\\\\sf\dashrightarrow\boxed{\pink{\sf{ KE = 18706 \ J }}}$

• Hence the Kinetic energy is 18706 Joules or 18.706 kJ .

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• Question 49 :

$\red{\frak{Given}}\begin{cases}\sf The \ CO_2 \ gas . \\\sf Temperature = 27^{\circ}= 300K \end{cases}$

The square root of the average velocity squared in the molecules of a gas is called RMS velocity The RMS velocity can be calculated by ,

$\dashrightarrow\red{\sf V_{(RMS)}= \sqrt{\dfrac{3RT}{M}} }$

• Here R is the constant whose value is 8.314 J / mol.K . M is the molecular mass. On substituting the respective values ,

$\sf\dashrightarrow v_{(RMS)}= \sqrt{ \dfrac{ 3\times 8.314 \times 10^7\times 300}{44}}\\\\\sf\dashrightarrow v_{(RMS)}= \sqrt{\dfrac{7482.6\times 10^7}{44} }\\\\\sf\dashrightarrow v_{(RMS)}= \sqrt{ 170.059 \times 10^7 } \\\\\sf\dashrightarrow \boxed{\pink{\sf v_{(RMS)}= 41000 \ cm/s }}$

• Hence the RMS velocity is 4.1 × 10⁴ cm/s .

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