
A gas mixture consists of $ 3\cdot 0 $ mole of hydrogen and $ 2\cdot 0 $ mole of helium at absolute temperature T considering all vibrational modes (assuming internal energy at T=0), the total energy of system is
(A) $ 11\cdot 5 $ RT
(B) $ \text{1}0\cdot \text{5} $ RT
(C) $ \text{13}\cdot 5 $ RT
(D) 15 RT
Answer
554.1k+ views
Hint: We have given Monoatomic gas helium and diatomic gas hydrogen. Use the following formula: specific heat capacity at constant volume for monatomic gas is $ \dfrac{3}{2}R $ .
Specific heat capacity at constant volume for diatomic gas is $ $ $ \dfrac{5}{2}R $
The total energy of the system is given by.
$ {{n}_{1}}{{c}_{1v}}T+{{n}_{2}}{{c}_{2v}}T $
Here, n1 is no. of moles for one gas [hydrogen],
$ {{C}_{1v}} $ is specific heat capacity [hydrogen],
$ {{n}_{2}} $ is no. of moles for helium gas ,
$ {{C}_{2v}} $ is specific heat capacity for helium gas.
Complete step by step solution
We have given, Hydrogen gas which is diatomic. $ {{C}_{1v}} $ specific heat capacity at constant volume,
$ \dfrac{5}{2}R $ .
Here, R is gas constant.
No, of moles of hydrogen gas $ \left( {{n}_{1}} \right)=3 $
Now, for helium gas which is monoatomic. $ {{C}_{2v}} $ specific heat capacity for helium gas at constant volume is given by, $ \dfrac{3}{2}R $ .
No. of moles for helium gas = 2
Now, both the gases are mixed, then total internal energy of system is given by,
$ {{n}_{1}}{{c}_{1v}}T+{{n}_{2}}{{c}_{2v}}T $ ---------- (1)
Put all the above values in eq. (1).
$ \left( 3 \right)\left( \dfrac{5}{2}R \right)T+2\left( \dfrac{3}{2}R \right)T $
$ \dfrac{15}{2}RT+\dfrac{6}{2}RT $
$ \left( \dfrac{15+6}{2} \right)RT=\dfrac{21}{2}RT=10.5RT $
Hence, total energy of the system is $ 10\cdot 5 $ RT.
Note
Monoatomic gases consist of only one atom, it is capable of translatory motion and hence 3 degrees of freedom. Diatomic gases like hydrogen, Nitrogen, Oxygen, etc. have two atoms in it. The molecule is capable of translatory motion of its centre of mass as well as the molecule can rotate about its centre of mass. Therefore, a diatomic molecule has in all 5 degrees of freedom, three due to translation and two due to rotation.
Specific heat capacity at constant volume for diatomic gas is $ $ $ \dfrac{5}{2}R $
The total energy of the system is given by.
$ {{n}_{1}}{{c}_{1v}}T+{{n}_{2}}{{c}_{2v}}T $
Here, n1 is no. of moles for one gas [hydrogen],
$ {{C}_{1v}} $ is specific heat capacity [hydrogen],
$ {{n}_{2}} $ is no. of moles for helium gas ,
$ {{C}_{2v}} $ is specific heat capacity for helium gas.
Complete step by step solution
We have given, Hydrogen gas which is diatomic. $ {{C}_{1v}} $ specific heat capacity at constant volume,
$ \dfrac{5}{2}R $ .
Here, R is gas constant.
No, of moles of hydrogen gas $ \left( {{n}_{1}} \right)=3 $
Now, for helium gas which is monoatomic. $ {{C}_{2v}} $ specific heat capacity for helium gas at constant volume is given by, $ \dfrac{3}{2}R $ .
No. of moles for helium gas = 2
Now, both the gases are mixed, then total internal energy of system is given by,
$ {{n}_{1}}{{c}_{1v}}T+{{n}_{2}}{{c}_{2v}}T $ ---------- (1)
Put all the above values in eq. (1).
$ \left( 3 \right)\left( \dfrac{5}{2}R \right)T+2\left( \dfrac{3}{2}R \right)T $
$ \dfrac{15}{2}RT+\dfrac{6}{2}RT $
$ \left( \dfrac{15+6}{2} \right)RT=\dfrac{21}{2}RT=10.5RT $
Hence, total energy of the system is $ 10\cdot 5 $ RT.
Note
Monoatomic gases consist of only one atom, it is capable of translatory motion and hence 3 degrees of freedom. Diatomic gases like hydrogen, Nitrogen, Oxygen, etc. have two atoms in it. The molecule is capable of translatory motion of its centre of mass as well as the molecule can rotate about its centre of mass. Therefore, a diatomic molecule has in all 5 degrees of freedom, three due to translation and two due to rotation.
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