A sample of nitrogen gas occupies 1.55L at 27C and 1.00atm pressure What will the volume be at -100C?



Answer:
Charle's Law. Direct proportion. (Pressure constant)
V1 x T2 = V2 x T1
1.55L x -100°C (173K) = V2 x 27°C (300K)
173 ÷ 300 = 0.577
1.55L x 0.577 = 0.894L = 894mL Final volume.
Assuming that the pressure remains the same, the volume will be 1.55 x 173/300 liters.
27C + 273.15 = 300.15K (kelvin)

-100C + 273.15 = 173.15K

P1V1/T1 = P2V2/T2

I'll assume pressure is constant since it wasn't mentioned.

V1/T1 = V2/T2

V1T2/T1 = V2

(1.55L)(173.15K)/(300.15K) = V2

0.894L = V2
V1/V2=T1/T2
1.55L/V=(27+273)K / ( -100+273) K
1.55L/V = 300K / 173 K

V=1.55 L * 173 /300
= 0.894L
u should use t as k not as celsius

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