What is the pressure of 380 mm Hg column of a gas in pascal ?
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Helium gas is compressed to half of the volume at 303 K. It should be heated to which temperature for its volume to increase to double of its original volume ?
When a gas is heated from 298K to 323K at a constant pressure of 1 atm its volume is
Hint 1 $ 380mm = 380 torr = { 380 \over 760 } atm $ $ \therefore 0.987 atm = 10 ^ 5 Pa $ $ \therefore { 380 \over 760 } atm = (?) \therefore 380 mm = 5.05 \times 10 ^ 4 Pa $ Hint 2 At const pressure according to charles law $ { V_1 \over T_1 } = { V_2 \over T_2 } = K $ $ \therefore { 22.4 \over 273} = 0.082 ......(1) and { 30.6 \over 373 } = 0.082 ---(2) $ Hint 3 According to Charles’ law $ { V_1 \over T_1 } = { V_2 \over T_2 } = K $ $ V_1 = { V_1 \over 2 } at T_1 = 303 K $ $ \therefore T_2 = { V_2 T_1 \over V_1 } $ $ V_2 = 2 V_1 at T_2 = ? $ $ = { 2 V_1 \times 303 \over {V_1 \over2 } } = 303 \times 4 = 1212 K $ Hint 4 According to Charles’ law $ { V_1 \over T_1 } = { V_2 \over T_2 } $ $ { V_1 \over 298 } = { V_2 \over 323} $ $ \therefore V_2 = { 323 \over 298 } V_1 = 1.08 V_1 $
At $ 20 ^\circ and 760 torr, the sample of air contains 20 \% O_2 \& 80 \% N_2 gaseous mixture, find the density of the air (Molarmass O_2 = 32 g/mol , N_2 = 28 g/mole R =0.082 liter mole ^ {-1} K ^ {-1}) $
The ratio of velocities of diffusion of gas A and B is 1 : 4, if the ratio of their masses in a mixtare is 2:3, calculate the ratio of their mole fractions (BIT 1990)
Under same conditions of temperature and pressure the volumes of $ 14g N_2 and 36g of O_3 $ are related as :
Equal masses of Hydrogen and oxygen gases are placed in a closed container, at a pressure of 3.4 atm. The contribution of hydrogen gas to the total pressure is
The kinetic energy of 4.0 moles of $ N_2 gas at 127 ^\circ C is (R =2 cal mole ^ {-1} K^{-1}) $
The critical temperature of $ H_2O is higher than CO_2 $ because (IIT 1997)
Critical temperature is directly proportional to dipole moment
The compressibility factor of an ideal gas is
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