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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 } $
The pressure and temperature of two different gases P and T having the volumes V for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be,
When two gases with the same volume and temperature are mixed, their pressures add up. Since both gases have the same pressure P, the total pressure of the mixture will be 2P.
Air is filled at $ 60 ^\circ C$ in a vessel of open mouth. The vessel is heated to a temperature T so that 1/4 th part of air escapes. Assuming the volume of the vessel remaining constant the value of T is.
For open mouth vessel, pressure is constant. volume is also given constant. Hence from $ \therefore PV = RT $ $ PV = { M \over M_o } RT \Rightarrow T \alpha { 1 \over M } $ $ \therefore { T_1 \over T_2 } = {M_2 \over M_1 } $ 1/ 4 th part escapes, so remaining mass in the vessel is $ M_2 = {3 \over 4 } M_1 \Rightarrow { 273 + 60 \over T } = { { 3 \over 4 } M_1 \over M_1 } \Rightarrow T = 444 K = 171 C $
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