i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease the temperature to 230 k, and measure the final volume to be 48 liters, what was the initial volume of the gas
Assuming that the number of mols are constant for both conditions: [tex] \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex] Now you plug in the given values. V_1 is the unknown. [tex]\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}
[/tex] Separate V_1 [tex]V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }[/tex] V= 162.782608696 L
