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QJEEM11Ph5f7a: JEE Main 2011 Physics - Class 11 JEEM11Ph5f7a Solution: Three perfect gases at absolute temperatures and are mixed. The …

Three perfect gases at absolute temperatures T1, T2{T_1},\,{T_2} and T3{T_3} are mixed. The masses of molecules are m1,m2{m_1},{m_2} and m3{m_3} and the number of molecules are n1,{n_1}, n2{n_2} and n3{n_3} respectively. Assuming no loss of energy, the final temperature of the mixture is:

Multiple Choice Options

  • A n1T1+n2T2+n3T3n1+n2+n3{{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}
  • B n1T12+n2T22+n3T32n1T1+n2T2+n3T3{{{n_1}T_1^2 + {n_2}T_2^2 + {n_3}T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}
  • C n12T12+n22T22+n32T32n1T1+n2T2+n3T3{{n_1^2T_1^2 + n_2^2T_2^2 + n_3^2T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}
  • D (T1+T2+T3)3{{\left( {{T_1} + {T_2} + {T_3}} \right)} \over 3}

Step-by-Step Verified Solution

Number of moles of first gas =n1NA= {{{n_1}} \over {{N_A}}}
Number of moles of second gas =n2NA= {{{n_2}} \over {{N_A}}}
Number of moles of third gas =n3NA= {{{n_3}} \over {{N_A}}}
If there is no loss of energy then
P1V1+P2V2+P3V3=PV{P_1}{V_1} + {P_2}{V_2} + {P_3}{V_3} = PV
n1NART1+n2NART2+n3NART3{{{n_1}} \over {{N_A}}}R{T_1} + {{{n_2}} \over {{N_A}}}R{T_2} + {{{n_3}} \over {{N_A}}}R{T_3}
=n1+n2+n3NARTmix= {{{n_1} + {n_2} + {n_3}} \over {{N_A}}}R{T_{mix}}
⇒Tmix=n1T1+n2T2+n3T3n1+n2+n3\Rightarrow {T_{mix}} = {{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}

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