You have a sample of argon gas at a certain pressure, volume, and temperature. You double the volume, double the number of moles of argon, and double the Kelvin temperature. How does the final pressure (P_f) compare to the original pressure (P_o)? P_f = A) (1/8)P_o B) (1/2)P_o C) 2P_o D) 4P_o E) 8P₂

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Quizplus · Accepted Answer

Using the ideal gas law, $PV = nRT$, if you double the volume ($V$), double the number of moles ($n$), and double the temperature ($T$), the pressure ($P$) remains unchanged because the changes in $n$ and $T$ increase the pressure, but the change in $V$ decreases it. However, since both $n$ and $T$ are doubled, their effect on pressure is quadrupled ($2 \times 2 = 4$), but doubling the volume halves the pressure. Thus, $4 \times 1/2 = 2$, so the final pressure is twice the original pressure.

You Have a Sample of Argon Gas at a Certain