Question

A vessel of volume V0 contains an ideal gas at pressure p0 and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals 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.

Answer 1

In 2 = rt/Vo

t = In^2ro/r

Explanation:

dV/ dt = r  

It implies dV = r

Let the pumped out gas pressure be dp.

Volume of container = V0

At a pump, dv amount of gas has been pumped out.

Pdv = –V0df  

Pv df = –V0 dp

Differential applied on both sides, we get:

Differential of P to P of dp/p = differential of 0 to 5 of dtr/Vo

We get P = P e^-rt/Vo

Please refer the attached picture for corret formatting of the equation.

Half of the gas has been pumped out.

Pressure will be half = 1/2e^-vt/vo

In 2 = rt/Vo

t = In^2ro/r

Answer 2

(a) The pressure of a gas $P=P_{0} e^{\frac{-r t}{V_{0}}}$

(b) The time taken $t=\frac{V_{0} \ln 2}{r}$

Explanation:

Let P be the pressure and n be the number of gas moles at any given time t inside the vessel.  

Suppose a small amount of dn moles gas is pumped out, and the pressure drop is dP.  

If we apply state equation to the gas inside the vessel, we obtain

$\begin{array}{l}(P-d P) V_{O}=(n-d n) R T \\P V_{O}-d P V_{O}=n R T-d n R T\end{array}$

But  

$P V_{o}=n R T$

$V_{o} d P=d n R T$     ……….(1)

The pressure carried out by the gas is equal to the pressure inside.  

Applying state equation, we get

$(P-d P) d V=d n R T$

$P d V=d n R T$  ……………( 2)

By equation 1 & equation 2, We've got

 $\begin{aligned}&V_{o} d P=P d V\\&\frac{d P}{P}=\frac{d V}{V_{o}}\end{aligned}$

$\begin{aligned}&\frac{d V}{d t}=r\\&d V=r d t\end{aligned}$            

$d V=-r d t$               ……………..(3)

As the pressure drops, the rate is negative

Now,

 $\frac{d P}{P}=-\frac{r d t}{V_{o}}$    (From equation 3)              

(a) Integrating the above  equation from $P_0$ to P and time o to t , we get

$\int_{P_{0}}^{P} \frac{d P}{P}=\int_{0}^{t}-\frac{r d t}{V_{o}}$

$\ln P-\ln P o=-r t V o$

$\ln \left(\frac{P}{P_{O}}\right)=-\frac{r t}{V_{o}}$

$P=P_{0} e^{\frac{-r t}{V_{0}}}$

(b)

$P=\frac{P_{0}}{2}$

$\frac{P_{0}}{2}=P_{0} e^{\frac{-r t}{V_{0}}}$

$e^{\frac{r t}{V_{o}}}=2$

$\frac{r t}{V_{o}}=\ln 2$

$t=\frac{V_{0} \ln 2}{r}$

Therefore the pressure of the gas as a function of time is $P=P_{0} e^{\frac{-r t}{V_{0}}}$ and the time taken before half the original gas is pumped out is $t=\frac{V_{0} \ln 2}{r}$