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The temperature of a gas at pressure P and volume V is $ 27 ^\circ C$ Keeping its volume constant if its temperature is raised to $ 927 ^\circ C$ , then its pressure will be
using Gay - Lussac ' law $ { P_1 \over P_2} = { T_1 \over T_2} $
Air is filled in a bottle at atmospheric pressure and it is corked at $ 35 ^\circ C$ , If the cork can come out at 3 atmospheric pressure then upto what temperature should the bottle be heated in order to remove the cork
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 what temperature volume of an ideal gas becomes triple
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 $
To double the volume of a given mass at an ideal gas at $ 27 ^\circ C$ keeping the pressure constant one must raise the temperature in degree centigrade
$ V \alpha T \Rightarrow { V_1 \over V_2 } = {T_1 \over T_2 } $
At constant temperature on incerasing the pressure of a gas 5% its volume will decrease by
$ 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 $
Hydrogen gas is filled in a ballon at $ 20 ^\circ C$ . If temperature is made $ 40 ^\circ C$, pressure remaining the same what fraction of hydrogen will come out
$ 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 $
When the pressure on 1200 ml of a gas is increased from 70 cm to 120 cm of mercury at constant temperature, the new volume of the gas will be
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 } $
A gas at $ 27 ^\circ C$ C has a volume V and pressure P. On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be
$ { 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 $
A perfect gas at $ 27 ^\circ C$ is heated at constant pressure to $ 327^\circ C$ . If original volume of gas at $ 27 ^\circ C$ is V then volume at $ 327 ^\circ C$ is
$ V \alpha T \Rightarrow { V_1 \over V_2 } = { T_1 \over T_2 } $
A vessel contains 1 mole of $O_2$ gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at temperature 2T has a pressure of
$ PV = RT \Rightarrow P = T $ ( V and R = constant ) $ \Rightarrow { P_2 \over P_1 } = { T_2 \over T_1 } $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.