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Physics NEET Questions — Free MCQ Practice

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4.0 gm ideal gas is filled in a bulb having volme $ 10 dm^3 at a constant temperature T & constant pressure P. If 0.8 gm gas is removed from the bulb to maintain the original pressure at (T + 125)K temperature, what would be the value of T for a gas having molar mass 40 gm mole ^{-1} $ .

A. 500K B. $ 500 ^\circ C $ C. 773 K D. $ 773 ^\circ C $

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Explanation

$ So : n_1T_1 = n_2T_2 $ $ {W_1 \over M_1 } \times T_1 = { W_2 \over M_2} T_2 M_1 = M_2 $

$ W_1T_1 = 3.2(T +125) $ 4T= 3.2T + 400 $ \therefore 0.8T = 400 $ $ \therefore T = 500K

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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 ?

A. 303K B. 606K C. 1212K D. $ 30 ^ \circ C $

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When a gas is heated from 298K to 323K at a constant pressure of 1 atm its volume is

A. Increases from V to 1.8 V B. Increases from V to 1.08 V C. Increases from V to 1.5 V D. Increases from V to 2 V

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Explanation

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 Charlesâ€™ 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 Charlesâ€™ 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 $

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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}) $

A. $ 1.918 gL ^ {-1} $ B. $ 2.198 gL ^ {-1} $ C. $ 1.198 gL ^ {-1} $ D. $ 1.394 gL^{-1} $

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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)

A. 1:12 B. 1:24 C. 1:6 D. 4 :3

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Under same conditions of temperature and pressure the volumes of $ 14g N_2 and 36g of O_3 $ are related as :

A. $ 2 V_{N_2} = 3 V_{o_3} $ B. $ 3 V_{N_2} = 2 V_{o_3} $ C. $ 3 V_{N_2} = 4 V_{o_3} $ D. $ 4 V_{N_2} = 3 V_{o_3} $

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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

A. 1.7 atm B. 0.2 atm C. 3.2 atm D. 3.02 atm

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The kinetic energy of 4.0 moles of $ N_2 gas at 127 ^\circ C is (R =2 cal mole ^ {-1} K^{-1}) $

A. 4400 cal B. 3200 cal C. 4800 cal D. 1524 cal

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The critical temperature of $ H_2O is higher than CO_2 $ because (IIT 1997)

A.

fewer electrons than $CO_2 $

B.

It contains 2- covalent bonds

C.

Molecular shape is V- shape

D.

dipole moment of $H_2O $

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Explanation

Critical temperature is directly proportional to dipole moment

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The compressibility factor of an ideal gas is

A. zero B. 1 C. 2 D. 4

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Physics MCQs for NEET — Practice Questions with Answers

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