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What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and
273 K. $ (R = 8.314 Jmol k^{-1})$
$ PV = RT \Rightarrow = {PV \over RT } $ $ Mass of \mu litre nitrogen = \times M_o $
A vessel contains 1 mole of $O_2$ gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing 1 mole of He gas (relative molar mass 4) at a temperature 2T has pressure of ..........
$PV = RT \Rightarrow P = { RT \over V} $ $ For same , R and V, P \alpha T $
The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be
molecular weight of oxygen = 32 (g) number of moles in 5 g of oxygen = 5 /32 $ \therefore Equation of state is PV = RT$ $ PV = { 5 \over 32 } RT$
A gas at 1 atmosphere and having volume 100 ml is mixed with another gas of equal moles at 0.5 atm and having volume 50 ml in flask of one litre, what is the final pressure?
Total number of moles is conserved, $ \therefore { P_1 V_1 \over RT} + { P_2 V_2 \over RT } = { PV \over RT} $ $ \therefore { 1 \times 100 \over RT } + { 0.5 \times 50 \over RT} = { P \times 1000 \over RT} $ ( 1 L = 1000 ml ) P = 0.125 atm
The quantity $ { PV \over k_E T } $ represents
The ideal gas equation is 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 $
Equation of gas in terms of pressure (P), absolute temperature (T) and density ($ \rho $) is
PV = RT $ \therefore Pv = \left( { M \over M_0 } \right) RT $ $ = \left( { M \over V } \right) {RT \over M_o } $ $ { \rho RT \over M_o } $ $ \left( {M \over V } = \rho \right) $ $ \Rightarrow { P \over \rho T } = { R \over M_o } = const $
$O_2$ gas is filled in a vessel. If pressure is doubled, temperature becomes four times, how many times its density will become.
PV = RT $ \therefore p \left( { M \over \rho} \right) = RT $ $ \therefore P \left( {M \over \rho} \right) = \left( { M \over M_o} \right) RT \Rightarrow \alpha { P \over T } $
At a given volume and temperature the pressure of a gas
PV = RT $ \therefore PV = \left( {M \over M_o} \right) RT \Rightarrow \rho \alpha { P \over T } $ ( V,T --> constant )
To decrease the volume of a gas by 5% at constant temperature the pressure should be
PV = RT = constant ( temp is const ) $ P_1 V_1 = P_2 V_2 $ $ P_1 V_1 = P_2 \left( { 95 \over 100} \right) V_1 $ $ ( V_2 = 95 \% V_1 ) $ $ \therefore P_2 = 1.0526 P_1 $ $ =P_1 + 0.0526 P_1 $ $ = P_1 + 5.26 \% P_1 $ $ \therefore Pressure 5.26 \% increases $
A gas at the temperature 250 K is contained in a closed vessel. If the gas is heated through 1 K, then the percentage increase in its pressure will be
Closed vessel i.e volume remains constant From PV = RT $ P \alpha T \therefore {P_2 \over P_1 } = { T_2 \over T_1} $ $ \therefore { P_2 - P_1 \over P_1 } = { T_2 - T_1 \over T_1 } $
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