You have a balloon containing 3 L of air at STP in a vacuum chamber. What will the volume of the balloon be when you reduce the pressure to 50 kPa and increase the temperature to 450 K?
A) 7.00 L
B) 9.89 L
C) 0.68 L
D) 5.46 L
we can use the combined gas law equation to find volume of gas [tex] \frac{P1V1}{T1} = \frac{P2V2}{T2} [/tex] P - pressure V - volume T - temperature parameters at STP conditions are on the left side and parameters for the second instance are on the right side of the equation standard temperature is 273 K and standard pressure is 10⁵ Pa substituting the values in the equation [tex] \frac{ 10^{5}Pa*3L }{273 K} = \frac{50000Pa*V}{450K} [/tex] V = 9.89 L new volume is 9.89 L answer is B) 9.89 L
