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ChemistryJAMB UTME2018

Options:

A.
B.
C.
D.
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Earlier discussion

3 older comments kept for context.

Favour Onwuchekwa

I agree with you on that answer.
Initial pressure=\( P_{1} \)
Initial volume=\( V_1 \)
Initial Temperature=\( T_1 \)=273K
Final pressure=Double the Initial pressure=2\( P_1 \)
Final volume=Double the Initial volume=2\( V_1 \)
Final Temperature=\( T_2 \)
Using the general gas equation
\( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \)
\( \frac{P_1V_1}{273}=\frac{2P_1×2V_1}{T_2} \)
Cross multiplying, we have;
\( P_1V_1×T_2=2P_1×2V_1×273 \)
\( P_1V_1T_2=1092P_1V_1 \)
\( P_1V_1 \) are on both sides therefore they'll cross out each other.
\( \therefore P_2=1092K \)
Option [D] is 💯% correct.

Favour Onwuchekwa

A little correction is to be made \( \therefore T_2 =1092K \)

Deborah Oludele

Option D