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Question: Two monatomic ideal gas at temperature \(T_{1}\) and \(T_{2}\) are mixed. There is no loss of energy...

Two monatomic ideal gas at temperature T1T_{1} and T2T_{2} are mixed. There is no loss of energy. If the masses of molecules of the two gases are m1m_{1} and m2m_{2} the number of their molecules are n1n_{1} and n2n_{2} respectively. The temperature of the mixture will be:

& A.\dfrac{{{T}_{1}}+{{T}_{2}}}{{{n}_{1}}+{{n}_{2}}} \\\ & .B.\dfrac{{{T}_{1}}}{{{n}_{1}}}+\dfrac{{{T}_{2}}}{{{n}_{2}}} \\\ & C.\dfrac{{{n}_{2}}{{T}_{1}}+{{n}_{1}}{{T}_{2}}}{{{n}_{1}}+{{n}_{2}}} \\\ & D.\dfrac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}}{{{n}_{1}}+{{n}_{2}}} \\\ \end{aligned}$$
Explanation

Solution

Ideal gas law or the general gas equation is the combination of Boyles Law, Charles’s law, Avogadro’s law and Gay Lussac’s law. It gives the relationship between the pressure PP applied on a VV volume of the gas which contains nn number of molecules at temperature TT .

Formula used:
PV=nRTPV=nRT

Complete step by step answer:
Let us consider an ideal gas. Then we can say that an isothermal process is where the system undergoes very slow changes like expansion or compression of the gas to avoid the loss of heat.
Let us assume the temperature of the two ideas gases to be T1T_{1} and T2T_{2} with mass m1m_{1} and m2m_{2} and n1n_{1} and n2n_{2} number of their molecules respectively. Let Cv1C_{v1} and Cv2C_{v2} be the internal energy of the gases at constant volume.
Given that, there is no loss of energy, then from the first law of thermodynamics, we can say that, n1Cv1ΔT1+n2Cv2ΔT2=0n_{1}C_{v1}\Delta T_{1}+n_{2}C_{v2}\Delta T_{2}=0
Since the gases are monatomic, we can say that Cv1=Cv2=Cv=32RC_{v1}=C_{v2}=C_{v}=\dfrac{3}{2}R
Let us also assume that the final temperature is TT, then,
32R[n1(TT1+n2(TT2)]=0\dfrac{3}{2}R[n_{1}(T-T_{1}+n_{2}(T-T_{2})]=0
    T=n1T1+n2T2n1+n2\implies T=\dfrac{n_{1}T_{1}+n_{2}T_{2}}{n_{1}+n_{2}}
Hence the correct option is D.n1T1+n2T2n1+n2D.\dfrac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}}{{{n}_{1}}+{{n}_{2}}}

Additional information:
However, the ideal gas law doesn’t give any information of the nature of reaction, i.e. when the gas is expanding or compressing does it absorb heat or release heat. Also as the name suggests these gases are ideal and such gases don't exist in the real world they are hypothetical in nature.

Note:
From ideal gas law, we know that PV=nRTPV=nRT where PP is the pressure applied on the and VV is the volume of the gas which contains nn number of molecules at temperature TT and RRis the gas constant. We can vary the different parameters to understand the behaviours of the gas in various conditions.