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JAMB Physics 2012 Paper

Question : 16

Total: 42

Solution:

Concept:

The combined gas law relates the pressure, volume, and temperature of a fixed mass of gas.

Explanation:

Standard temperature and pressure (s.t.p.) are

T_2 = 273 \, \mathrm{K}

and

P_2 = 760 \, \mathrm{mmHg}

Convert the initial temperature to kelvin:

T_1 = 27^\circ\mathrm{C} + 273 = 300 \, \mathrm{K}

Use the combined gas equation:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Make

V_2

the subject:

V_2 = \frac{P_1V_1T_2}{P_2T_1}

Substitute the values:

V_2 = \frac{700 \times 2000 \times 273}{760 \times 300}

Simplify:

V_2 = \frac{382200000}{228000} = 1676.3 \, \mathrm{cm^3}

So the volume at standard temperature and pressure is

1676.3 \, \mathrm{cm^3}

Answer:

1676.3 \, \mathrm{cm^3}

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