Question

A gas originally occupying 10.1 L at 0.925 atm and 25 C is changed to 12.2 L at 625 torr. What is the new temperature?

Answer 1

We can use the combined gas law equation to find the new temperature 
$\frac{P1V1}{T1} = \frac{P2V2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
P - pressure 
P1 - 0.925 atm x 101 325 Pa/atm = 93 726 Pa
P2 - 625 torr x 133 Pa/torr = 83 125 Pa
V - volume 
V1 - 10.1 L
V2 - 12.2 L
T - temperature in kelvin
T1 -  25 °C + 273 = 298 K
substituting these values in the equation
$\frac{93726Pa * 10.1L }{298K} = \frac{83125 Pa*12.2L}{T2}$
T = 319 K
temperature in celsius - 319 K - 273 = 46 °C