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JAMB Physics 2021 Paper

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Question : 7

Total: 40

A given mass of gas has a pressure of 80 Nm

^{-2}

at a temperature of 47 C. If the temperature is reduced to 27 C with volume remaining constant, the new pressure is?

46.0 Nm $^{-2}$
75.0 Nm $^{-2}$
80.0 Nm $^{-2}$
85.3 Nm $^{-2}$

Solution:

Concept:

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

Explanation:

Use the pressure law:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Convert all temperatures from Celsius to Kelvin.

T_1 = 47 + 273 = 320\, \mathrm{K}

T_2 = 27 + 273 = 300\, \mathrm{K}

Given

P_1 = 80\, \mathrm{N\,m^{-2}}

.

Make

P_2

the subject:

P_2 = \frac{P_1 T_2}{T_1}

Substitute the values:

P_2 = \frac{80 \times 300}{320} = 75\, \mathrm{N\,m^{-2}}

Answer:

The new pressure is

75.0\, \mathrm{N\,m^{-2}}

, which is option B.

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