JAMB - Chemistry (1985 - No. 44)
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?
18.6k
100.0k
298.0k
1192.0k
4768.0k
Wyjaśnienie
\(\frac{P_1V_1}{T_1} = \frac{4P_1 \times 4V_1}{T_2}\)
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
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