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?

  • A 18.6k
  • B 100.0k
  • C 298.0k
  • D 1192.0k
  • E 4768.0k
jamb 1985chemistry

The correct answer is E. 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

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