Calculate the temperature of 6 moles of an ideal gas at a pressure of 7.6 x 106Nm-2 with a volume of 10-3m3.
[R = 8.3jmol-1K-1]
[R = 8.3jmol-1K-1]
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Correct Answer: Option C
Explanation:
T = \(\frac{PV}{nR}\) = \(\frac{7.6 x 10^6 x 10^-3}{6 x 8.3}\)
= 153oC
T = \(\frac{PV}{nR}\) = \(\frac{7.6 x 10^6 x 10^-3}{6 x 8.3}\)
= 153oC