Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

$ VP^3 = constant = k \Rightarrow P^3 \alpha { 1 \over V} \Rightarrow P \alpha { 1 \over V^{1 /3 }} \Rightarrow P = {k \over V^{1/3} } $ $ PV = RT \Rightarrow { k \over V^{1/3 }}V = RT $ $ = k V^{2/3 } = RT $ $ = V^{2/3} = RT/ k $ $ Hence \left( { V_1 \over V_2 } \right)^{2/3} = { T_1 \over T_2} \Rightarrow \left( { V \over 27V } \right)^{2/3} = { T \over T_2 } $ $ \Rightarrow {1 \over 9 } = { T \over T_2} \Rightarrow T_2 = 9T $

using Gay - Lussac ' law $ { P_1 \over P_2} = { T_1 \over T_2} $

At constant volume $ { P_1 \over T_1 } = { P_2 \over T_2 } \Rightarrow T_2 = \left( P_2 \over P_1 \right) T_1 $

At constant pressure $ V \alpha T $ $ \Rightarrow { V_2 \over V_1 } = {T_2 \over T_1} \Rightarrow T_2 = \left( { V_2 \over V_1 } \right) T_1 $

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

$ 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 $

$ 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 $

At constant Pressure PV = constant $ \therefore P_1 V_1 = P_2 V_2 \Rightarrow {P_1 \over P_2} = { V_2 \over V_1 } $

$ { P_1 V_1 \over T_1 } = { P_2 V_2 \over T_2 } \Rightarrow T_2 = \left( { P_2 V_2 \over P_1 V_1 } \right) T_1 $

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