The density of helium gas at 0°C is $ \rho_0 = 0.179 \, \text{kg/m}^3 $. The temperature is then raised to $ T = 95°C $, but the pressure is kept constant. Assuming that helium is an ideal gas, calculate the new density $ \rho_f $ of the gas.[Input box for answer] kg/m$^3$ Need Help? [Button: Read It]

Helium is an ideal gas. The ideal gas equation is: Pv=nRTP=Absolute pressurev=Volumen=Number of…

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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The density of helium gas at 0°C is $ \rho_0 = 0.179 \, \text{kg/m}^3 $. The temperature is then raised to $ T = 95°C $, but the pressure is kept constant. Assuming that helium is an ideal gas, calculate the new density $ \rho_f $ of the gas.

[Input box for answer] kg/m$^3$

Need Help? [Button: Read It]

Expert Solution

Step 1

Helium is an ideal gas.

The ideal gas equation is:

$\begin{array}{rcl} P v & = & n R T \\ P & = & A b s o l u t e p r e s s u r e \\ v & = & V o l u m e \\ n & = & N u m b e r o f m o l e s \\ R & = & U n i v e r s a l G a s C o n s \tan t \\ T & = & A b s o l u t e T e m p e r a t u r e \end{array}$

Again, Number of moles,

$\begin{array}{rcl} n & = & \frac{ρ v}{M} \\ ρ & = & D e n s i t y o f g a s \\ M & = & M o l a r m a s s \\ v & = & v o l u m e o f t h e g a s \end{array}$

Step 2

Now, the temperature changes from $0^{\circ} C$ to $95^{\circ} C$

The pressure and molar mass remain constant. throughout the process.

At $0^{\circ} C$ the density is $ρ_{0} = 0.179 k g / m^{3}$

Now,

$\begin{array}{rcl} \frac{ρ_{f}}{ρ_{0}} & = & \frac{M \times n}{v_{f}} \times \frac{v_{0}}{M \times n} \\ = & \frac{v_{0}}{v_{f}} \\ = & \frac{n \times R \times T_{0}}{P} \times \frac{P}{n \times R \times T_{f}} \\ = & \frac{T_{0}}{T_{f}} \end{array}$

So, The density of helium gas at $95^{\circ} C$ is:

$\begin{array}{rcl} ρ_{f} & = & \frac{T_{0}}{T_{f}} \times ρ_{0} \\ = & \frac{(0 + 273.15) K}{(95 + 273.15) K} \times 0.197 \frac{k g}{m^{3}} \\ = & 0.146 \frac{k g}{m^{3}} \end{array}$

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