Physics MCQs for NEET — Practice Questions with Answers

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An ideal gas expands according to PV=constant. On expansion, the temperature of gas:

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Explanation

(c) Ideal gas do not show change in temperature during expansion.

Select the correct statement. In the gas equation PV = nRT                 

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Explanation

(c) The ideal gas equation is pV = nRT

where, V is the volume of n moles of a gas.

If a gas expands at constant temperature, it indicates that                                     

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Explanation

(c) KE =32RT (for one mole of a gas)

As, the kinetic energy of a gaseous molecule depends only on temperature, thus at constant temperature, the kinetic energy of the molecules remains the same.

The condition of SATP refers for:

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Explanation

(d) SATP means 1 bar and 25 °C .

Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm but also a concentration of 1 mol L-1? (R=0.082 L atm mol-1deg-1)                                     

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Explanation

(c) According to ideal gas equation,

                         pV = nRT

                          p = nVRT

                    1 atm = 1 mol L-1 x 0.082 x T

                          T =1/0.082 = 12 K

The compressibility of a gas is less than unity at STP. Therefore:

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Explanation

(b) The compressibility factor (Z) = (Px22.4)/RT = 1 (for ideal gas)

                                           (Z) = (P x Vm)/RT <1   (for real gas)

                                   ... 22.4/Vm > 1 or Vm < 22.4

The numberical value of cp -cv is equal to:

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Explanation

(b) Cp-Cv = R;

cp=M x Cp and cv = M x Cv

The van der Waals' equation for real gas is:

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Explanation

(d) These are van der Waals' equations for 1 mole (a) and n mole gas (b), (c).

How much should the pressure be increased in order to decrease the volume of a gas 5% at a constant temperature?

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Explanation

Applying Boyle's Law, P1V1=P2V2 

Suppose initial pressure is P and initial volume is V, then 

PV=P2 x 0.95V

P2=(p/0.95)=(100p/95)

Increase in pressure = (100P/95)-P=(5P/95)

% increase in pressure = {(5P/95)/P} x 100 = 500/95 = 5.26%

An open container is heated from 300 K to 400 K then % of gas remain in the container is:

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Explanation

For open container, 1/n1T1 = 1/n2T2 (P and V are constant)

Suppose initial moles of gas is n.

1/nx300 = 1/n2x400

n2 = (3/4)n

% of gas remain = (3/4)n/n x 100 = 75%

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