A sample of chlorine gas (Cl2) occupies 1.65 L at 1.0 atm and 308 K. What pressure (in mm Hg) will the sample exert at 3.00 L and 51.0°C?
I know the answer is 440 but I dont know how to get that answer so please show work.
This can be calculated using the ideal gas law pV = nRT
p1 V1 T2 = p2 V2 T1
51.0o C = 324 K
putting all the values in the expression we get ,
1.0 x 1.65 x 324 = p2 x 3.00 x 308
p2 = 0.578571 atm
1 atm = 760 mmHg
so 0.578571 atm = 0.578571 x 760
= 440 mmHg
Hope this helps !!
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