Question

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

Answer 1

Assuming that the number of mols are constant for both conditions:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Now you plug in the given values. V_1 is the unknown. 
$\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}$
Separate V_1
$V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }$
V= 162.782608696 L 

There are 2 sig figs

V= 160 L