Question

a gas has a volume of 2.00 at 323k and 3.00 ATM What would be the new volume if the temperature is changed to 273 K in the pressure is changed to 1ATM

Answer 1

Answer:

$V_2 = 5.07L$

Explanation:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\\V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{3.00atm \cdot 2.00L \cdot 273K}{323K \cdot 1.00atm} = 5.07L$