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The product of the pressure and volume of an ideal gas is
$ \ therefore PV \alpha T ( R \rightarrow constant ) $
At $ O ^\circ C$ the density of a fixed mass of a gas divided by pressure is x. At $ 100 ^\circ C $ , the ratio will be
$ \therefore PV \alpha T $ $ ( R --> constant ) $ $ PV = RT = \left( { M \over M_o } \right) RT $ $ \therefore { M \over PV } = { M_o \over RT} $ $ \therefore { density \over P } = { M_o \over RT } $ $ \therefore \left( { density \ over P } \right) _ { at 0 c } = { M \over R (273 ) } = x$ .......(i) $ \therefore \left( { density \ over P } \right) _ { at 100 c } = { M \over R (373 ) } $ .......(ii) $ \therefore \left( { density \ over P } \right) _ { at 100 c } = \left ( { 273 \over 373} \right) x$
Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the air temperature is $22 ^\circ C$ . During the day, temperature rises to $42 ^\circ C$ and the tube expands by 2% The pressure of the air in the tube at this temperature will be approximately.
$ { PV \over T } = R =constant \Rightarrow { P_1 V_1 \over V_1 } = {P_2 V_2 \over T_2 } $
The volume of a gas at $ 20 ^\circ C $ is 200 ml. If the temperature is reduced to $–20 ^\circ C$ at constant pressure, its volume will be.
PV=RT since P is const $ \therefore V \alpha T $ $ \Rightarrow { V_1 \over V_2 } = { T_1 \over T_2 } $
2g of $O_2$ gas is taken at $27 ^\circ C $ and pressure 76 cm Hg. Find out volume of gas (ln litre)
PV = RT $ = \left ( {M \over M_o} \right) RT \Rightarrow V = { MRT \over M_o P } $
1 mole of gas occupies a volume of 100 ml at 50 mm pressure. What is the volume occupied by two moles of gas at 100 mm pressure and at same temperature
PV =RT $ { P_1 V_1 \over P_2 V_2 } = {1 \over 2 } $ ( T constant )
A cylinder contains 10 kg of gas at pressure of $10^7 N/m^2$ . The quantity of gas taken out of the cylinder, if final pressure is $ 2.5 \times 10^ 6 Nm^{-2} $ , will be…………. (temperature of gas is constant)
PV = RT $ \therefore PV = \left( { M \over M_o} \right) RT \Rightarrow P \alpha M $ ( V,R,T - constant ) $ \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 $ Hence mass of the gas taken out of the cylinder = 10 – 2.5 = 7.5 kg
Suppose ideal gas equation follows $ VP^3 = constant$ , Initial temperature and volume of the gas are T and V respectively. If gas expand to 27 V, then temperature will become
$ 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 $
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 $
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