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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:

Using $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

P $_{1}$ = 80Nm $^{2}$

T

_{1}

= 47 + 273 = 320k

T

_{2}

= 27 + 273 = 300k

P

_{2}

?

$\therefore$ = P $_{2}$ = $\frac{P_{1}T_{2}}{T_{1}}$

= $\frac{80 \times 300}{320}$ = 75.Nm $^{2}$

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