Kinetic Theory of Gases NEET प्रश्न हिंदी में — भौतिक विज्ञान (Physics) MCQ उत्तर सहित

आदर्श गैस समीकरण है PV = RT PV = RT ( = 1 ) $ \ PV = k_B N_A T $ $ \left( { R \over N_A} = k_B \right) $ $ \therefore {PV \over k_B T } = N_A = Avogadro's number $

$PV = RT \Rightarrow P = { RT \over V} $ $ For same , R and V, P \alpha T $

$ P \alpha { 1 \over V } \Rightarrow { V_2 \over V_1 } = {P_1 \over P_2 } \Rightarrow { 100 \over 105 } \Rightarrow V_2 = { 100 \over 105 } V_1 = 0.9524V_1 $ $ \therefore V_2 = ( 1 - 0.0476) V_1 $ $ = V_1 - 0.476 V_1 $ $ = V_1 - 4.76 \% V_1 $

स्थिर तापमान पर PV = नियतांक $ \therefore P_1 V_1 = P_2 V_2 \Rightarrow {P_1 \over P_2} = { V_2 \over V_1 } $

PV = RT $ \therefore PV = \left( { M \over M_o} \right) RT \Rightarrow P \alpha M $ ( V,R,T - स्थिरांक ) $ \Rightarrow { P_1 \over P_2 } = { M_1 \over M_2 } \Rightarrow { 10^7 \over 2.5 \times 10^6} = { 10 \over M_2 } \Rightarrow M_2 = 2.5 kg $ अतः सिलेंडर से निकाली गई गैस का द्रव्यमान = 10 – 2.5 = 7.5 kg

$ V \alpha T \Rightarrow { V_2 \over V_1 } = {T_2 \over T_1 } \Rightarrow { V_2 - V_1 \over V_1 } = { T_2 - T_1 \over T_1 } $ $ \Rightarrow { \triangle V \over V } = { ( 273 + 40 ) - (273 + 20 ) \over 273 + 20 } = { 313 - 293 \over 293 } = 0.07 $

$ V _{rms } = \sqrt { 3P \over \rho} = \sqrt { 3PV \over M } \Rightarrow V_{rms } \alpha \sqrt { P \over M } $ $ \therefore{ V_1 \over V_2 }= \sqrt { {P_1 \over P_2 } \times { M_2 \over M_1 }} $

$ E \alpha T $ $ \therefore { E_1 \over E_2 } = { T_1 \over T_2 } $

$ V \alpha T \Rightarrow { V_1 \over V_2 } = { T_1 \over T_2 } $

$ \nu_{rms} = \sqrt { 3RT \over Mo } \Rightarrow \nu_{rms} \alpha \sqrt T $ $ \therefore T \alpha \nu_{rms} ^2 $