Name: Molar Specific Heat Capacities of Gases
Uploaded: 2019-07-19
Duration: 10 min 31 s
Description: Do all gases have the same molar heat capacities? Are there certain criteria for determining molar specific heat capacities of gases? Watch the video to find...

Question

Find the pressure of air in a vessel being evacuated as a function of evacuation time $t$ . The vessel's volume is $V$ and initial pressure is $p_{0} .$ The process is assumed to be isothermal and the evacuation rate equal to $C$ , and independent of pressure.

Note: The evacuation rate is the gas volume being evacuated per unit time, with that volume being measured under the gas pressure attained by that moment.

Accepted Answer

Using the ideal gas equation, we get,

$V$

Differentiating both sides with respect to time $t$ , we get,

$\frac{d p}{d t} = \frac{R T}{M V} \frac{d m}{d t} . . . (2)$

Let $V$ be the volume of the gas at time $t = t$ and suppose $∆ V$ be the volume is ejected in time $∆ t$ ,

$m (t) = m$ be the mass of the gas at time $t = t$ and

$m (t + ∆ t)$ be the mass of the gas at time $t = t + ∆ t$ . So, we have

$∆ V = \frac{V}{m (t + ∆ t)} [m (t) - m (t + ∆ t)] \Rightarrow \frac{∆ V}{∆ t} = \frac{V}{m (t + ∆ t)} \frac{[m (t) - m (t + ∆ t)]}{∆ t} \Rightarrow \lim_{∆ t \to 0} \frac{∆ V}{∆ t} = \lim_{∆ t \to 0} [- \frac{V}{m (t + ∆ t)} \frac{[m (t + ∆ t) - m (t)]}{∆ t}] \Rightarrow \frac{d V}{d t} = - \frac{V}{m} \frac{d m}{d t} . . . (3) [∵ \lim_{∆ t \to 0} m (t + ∆ t) = m (t) = m & \lim_{∆ t \to 0} \frac{[m (t + ∆ t) - m (t)]}{∆ t} = \frac{d m}{d t}]$

According to the question, the rate of evacuation $\frac{d V}{d t}$ is equal to $C$ . So, we have

$C = - \frac{V}{m} \frac{d m}{d t} \Rightarrow \frac{d m}{d t} = - \frac{C m}{V} . . . (4)$

Putting $\frac{d m}{d t}$ from equation $(4)$ into $(2)$ , we get,

$\frac{d p}{d t} = - \frac{C}{V} (\frac{m R T}{M V}) \Rightarrow \frac{d p}{d t} = - \frac{C}{V} p [using (1)]$

$\Rightarrow \frac{d p}{p} = - \frac{C}{V} d t$

Integrating within proper limits, we get,

$\int_{p_{0}}^{p} \frac{d p}{p} = - \frac{C}{V} \int_{0}^{t} d t$

$\Rightarrow \ln \frac{p}{p_{0}} = - \frac{C}{V} t \Rightarrow \frac{p}{p_{0}} = e^{\frac{- C t}{v}} \Rightarrow p = p_{0} e^{^{\frac{- C t}{v}}}$

Hence, pressure as a function of time is given by,

$p = p_{0} e^{^{\frac{- C t}{v}}}$ .

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