A certain volume of gas at 298k is heated such that its volume and pressure are now four times the original values. What is the new temperature?
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Correct Answer: Option E
Explanation:
\(\frac{P_1V_1}{T_1} = \frac{4P_1 \times 4V_1}{T_2}\)
i.e \(\frac{P_1V_1}{298} = \frac{4P_1 \times 4V_1}{T_2}\)
T2 = \(\frac{298 \times 4P_1 \times 4V_1}{P_1V_1}\)
= 298 + 16 = 4768.0k
\(\frac{P_1V_1}{T_1} = \frac{4P_1 \times 4V_1}{T_2}\)
i.e \(\frac{P_1V_1}{298} = \frac{4P_1 \times 4V_1}{T_2}\)
T2 = \(\frac{298 \times 4P_1 \times 4V_1}{P_1V_1}\)
= 298 + 16 = 4768.0k