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# A vessel of volume ${V_0}$contains an ideal gas at pressure ${P_0}$and temperature $T$. Gas is continuously pumped out of this vessel at a constant volume-rate $\dfrac{{dV}}{{dt}} = r$keeping the temperature constant, The pressure inside the vessel. Find (a) the pressure of the gas as a function of time,(b) the time taken before half the original gas is pumped out.

Last updated date: 17th Jun 2024
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Hint: It is given that an ideal gas is pumped out of a vessel, which has a volume ${V_0}$ at the constant rate of $\dfrac{{dV}}{{dt}} = r$. Use the ideal gas equation and apply the boundary conditions to find the relationship between pressure and time and use that to find the time taken before half the gas is out.

Let us assume that the volume of the vessel is ${V_0}$ and the ideal gas is at a pressure ${P_0}$at a temperature $T$. The initial conditions are given using ideal gas law equation which states,
$PV = nRT$
From this equation we know that Pressure and volume are constant. Hence on differentiating on both sides we get ,
$\Rightarrow PdV + VdP = 0$
From the given statement we know that,
$\dfrac{{dV}}{{dt}} = r$
Rearranging this , we get,
$dV = rdt$

Substituting this in equation , we get
$\Rightarrow P(r \times dt) + VdP = 0$
$\Rightarrow P(r \times dt) = - VdP$
Now, the gas is said to occupy the volume of the vessel, substituting this , we get
$\Rightarrow rdt = - {V_0}\dfrac{{dP}}{P}$
$\Rightarrow - \dfrac{r}{{{V_0}}}dt = \dfrac{{dP}}{P}$
Integrating on both sides , we get
$\Rightarrow - \dfrac{r}{{{V_0}}}\int {} dt = \int {\dfrac{{dP}}{P}}$
On applying laws of integration we get,
$\Rightarrow - \dfrac{r}{{{V_0}}}t + c = \ln (P)$-------(1)

Now, initially when the gas is inside the vessel, the pressure of the gas is ${P_0}$ and $t = 0$. Substituting this
$\Rightarrow - \dfrac{r}{{{V_0}}}(0) + c = \ln ({P_0})$
$\Rightarrow c = \ln ({P_0})$
Substituting c in equation 1 , we get,
$\Rightarrow - \dfrac{r}{{{V_0}}}t + \ln ({P_0}) = \ln (P)$
Removing ln on both sides , we get,
$\Rightarrow {e^{ - \dfrac{r}{{{V_0}}}t}} = \dfrac{P}{{{P_0}}}$ (using property $\ln (A) - \ln (B) = \ln (\dfrac{A}{B})$)
$\Rightarrow {e^{ - \dfrac{r}{{{V_0}}}t}}{P_0} = P$

Now, when half the volume of the gas is pumped out, we know that
$\Rightarrow \dfrac{{{P_0}{V_0}}}{{P{V_0}}} = \dfrac{{nRT}}{{\dfrac{{nRT}}{2}}}$, where the numerator represent the initial conditions and denominator represent final conditions when the gas is let out.
Re-arranging this , we get,
$\Rightarrow P = \dfrac{{{P_0}}}{2}$
Substituting this in the general pressure relation , we get,
$\Rightarrow {e^{ - \dfrac{r}{{{V_0}}}t}}{P_0} = \dfrac{{{P_0}}}{2}$
Cancelling out the common terms, we get
$\Rightarrow {e^{ - \dfrac{r}{{{V_0}}}t}} = \dfrac{1}{2}$
Removing the exponential term, we get,
$\Rightarrow - \dfrac{r}{{{V_0}}}t = \ln (\dfrac{1}{2})$
On rearranging this we get,
$\Rightarrow t = \dfrac{{{V_0}\ln 2}}{r}$
Thus the relation has been established between pressure as a function of time.

Note: Ideal gas is a theoretically made up gas which is composed of particles that don’t have intermolecular forces of attraction or repulsion between them under applied pressure and volume. This is considered to be the ideal state of the gas.