A helium-filled balloon has a volume of 100 L at 25°C and 900 mmHg. What volume will it have at 180 kPa and 20°C?

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A helium-filled balloon has a volume of 100 L at 25°C and 900 mmHg. What volume will it have at 180 kPa and 20°C?

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Answer 1 · 2016-06-30T08:33:17

The combined gas law holds that $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$

Explanation:

The temperatures are quoted in $degrees Kelvin$ $"degrees Kelvin"$ . Of course, the twist introduced is the pressure value. We know that $1 \cdot a t m$ $1*atm$ will support a column of mercury $760 \cdot m m$ $760*mm$ high, and the pressure measurement here is $900 \cdot m m$ $900*mm$ . This is not very realistic, as no one would measure those pressures with a mercury manometer. (Why not? Well. have you ever tried to clean up a mercury spill? Putting mercury under pressure is just asking for trouble.)

Thus $P_{1} = 900 \cdot m m \cdot H g = \frac{900 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{1}}$ $P_1=900*mm*Hg=(900*mm*Hg)/(760*mm*Hg*atm^1)$ $=$ $=$ $1.18 \cdot a t m$ $1.18*atm$

And $P_{2} = 180 \cdot k P a = \frac{180 \cdot k P a}{101.32 \cdot k P a \cdot a t m^{- 1}}$ $P_2=180*kPa=(180*kPa)/(101.32*kPa*atm^-1)$ $=$ $=$ $1.78 \cdot a t m$ $1.78*atm$

You now have consistent units, and I leave the calculation as an exercise.

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A helium-filled balloon has a volume of 100 L at 25°C and 900 mmHg. What volume will it have at 180 kPa and 20°C?

1 Answer

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