Question

Q15. What temperature (in °C) did an ideal gas shift to if it was initially at -11.0 °C at 4.33 bar and 35.0 L and the pressure was changed to 8.71 bar and the volume changed to 15.0 L?

Answer 1

$\frac{P_1×V_1}{T_1}= \frac{P_2×V_2}{T_2}$

$T_2=\frac{8.71×15×(273-11)}{4.33×35}=255.87=-47.13°c$