The average energy of molecules in a sample of oxgen gas at 300 K are 6.21 × 10-21 J. The corresponding values at 600 K are

Question

The average energy of molecules in a sample of oxgen gas at 300 K are 6.21 × 10-21 J. The corresponding values at 600 K are (A) 6.21 × 10-21 J (B) 8.78 × 10-21 J (C) 4.21 × 10-21 J (D) 12.42 × 10-21 J

Tardigrade Admin · Accepted Answer

The average kinetic energy of molecules in a sample of oxygen gas is given by

K E_{a vg} = \frac{3}{2} k T or \frac{( K E _{a vg} ) _{1}}{( K E _{a vg} ) _{2}} = \frac{T _{1}}{T _{2}}

(K E_{a vg})_{1} = 6.21 \times 1 0^{- 21} J, T_{1} = 300 K

T_{2} = 600 K, (K E_{a vg})_{2} = ?

∴ \frac{6.21 \times 1 0 ^{- 21}}{( K E _{a vg} ) _{2}} = \frac{300}{600} = \frac{1}{2}

(K E_{a vg})_{2} = 12.42 \times 1 0^{- 21} J

Q. The average energy of molecules in a sample of oxgen gas at $300 K$ are $6.21 \times 1 0^{- 21 J}$ . The corresponding values at $600 K$ are

$6.21 \times 1 0^{- 21} J$

$8.78 \times 1 0^{- 21} J$

$4.21 \times 1 0^{- 21} J$

$12.42 \times 1 0^{- 21} J$

Q. The average energy of molecules in a sample of oxgen gas at 300K are 6.21×10−21J. The corresponding values at 600K are

6.21×10−21J

8.78×10−21J

4.21×10−21J

12.42×10−21J

The average kinetic energy of molecules in a sample of oxygen gas is given by KEavg​=23​kTor(KEavg​)2​(KEavg​)1​​=T2​T1​​ (KEavg​)1​=6.21×10−21J,T1​=300K T2​=600K,(KEavg​)2​=? ∴(KEavg​)2​6.21×10−21​=600300​=21​ (KEavg​)2​=12.42×10−21J

Q. The average energy of molecules in a sample of oxgen gas at $300 K$ are $6.21 \times 1 0^{- 21 J}$ . The corresponding values at $600 K$ are

$6.21 \times 1 0^{- 21} J$

$8.78 \times 1 0^{- 21} J$

$4.21 \times 1 0^{- 21} J$

$12.42 \times 1 0^{- 21} J$