Question

Nitrogen gas is being withdrawn at the rate of 4.5 g/s from 0.15 m3 cylinder, initiallycontaining the gas at a pressure of 10 bar and 320 K. The cylinder does not conduct heat, nordoes its temperature change during the emptying process. What will be the temperature andpressure of the gas in the cylinder after 5 minutes? What will be the rate of change of the gastemperature at this time? Nitrogen can be considered to be and ideal gas with C*P = 30J/(mol K)

Answer 1

       $\text{From the mass balance equation we have:-}$

       $\dfrac{dN}{dt} = N$

$\implies \: \dfrac{V}{R}\dfrac{d\text{\huge{(}}\dfrac{P}{T}\text{\huge{)}}}{dt} = N$

$\implies \: \dfrac{d\text{\huge{(}}\dfrac{P}{T}\text{\huge{)}}}{dt} = N\dfrac{R}{V}$

$\implies \: \dfrac{-\dfrac{4.5}{28}\dfrac{mol}{s}\times8.314\dfrac{J}{molK}}{0.15\: m^3} = -8.908 \times 10^{-5} \dfrac{bar}{K.s}$

       $\text{Integrating this we get}$

$\implies \: \text{\huge{(}}\dfrac{P}{T}\text{\huge{)}}_{final} = \text{\huge{(}}\dfrac{P}{T}\text{\huge{)}}_{initial} - 8.908 \times 10^{-5}t$

$\implies \: \text{\huge{(}}\dfrac{P}{T}\text{\huge{)}}_{final} = \dfrac{10}{320} - 8.908 \times 10^{-5}t \text{ ------ (i)}$

       $\text{We know that:-}$

        $\text{\huge{(}}\dfrac{T_{final}}{T_{initial}}\text{\huge{)}}^{\frac{C_p}{R}} = \dfrac{P_{final}}{P_{initial}}$

$\implies \: \dfrac{T_{final}^{\frac{30}{8.314}}}{P_{final}} = \dfrac{T_{initial}^{\frac{30}{8.314}}}{P_{initial}} = \dfrac{320^{\frac{30}{8.314}}}{10} \text{ ------ (ii)}$

       $\text{From (i) and (ii)}$

$\implies \: T_{final} = 152.57 \; K \; ; \; P_{final} = 0.6907 \; bar$

       $\text{We have}$

       $\dfrac{C_v}{R}\dfrac{dlnT}{dt} = \dfrac{dln(\dfrac{P}{T})}{dt} = \dfrac{dlnP}{dt} - \dfrac{dlnT}{dt} = \dfrac{dP}{Pdt} - \dfrac{dlnT}{dt}$

$\implies \: \dfrac{C_v}{dt}\dfrac{dlnT}{dt} = \dfrac{T}{P}\text{\huge{\{}}\dfrac{1}{T}\text{\huge{(}}\dfrac{dP}{dt} - \dfrac{PdlnT}{dt}\text{\huge{)}}\text{\huge{\}}}$

$\implies \: \dfrac{T}{P}\dfrac{d(\dfrac{P}{T})}{dt} =\dfrac{NRT}{PV}$

$\implies \: \dfrac{C_v}{RT}\dfrac{dT}{dt} = \dfrac{NRT}{PV}$

$\implies \: \dfrac{dT}{dt} = \dfrac{N(RT)^2}{C_vPV}$

       $\text{at t = 5 min,}$

       $P = 0.6907 \: bar$

       $T = 152.57 \: K$

       $C_v = C_p - R = 30 - 8.314 = 21.686 \: J/molK$

       $\text{putting it back in the equation}$

$\implies \: \boxed{\dfrac{dT}{dt} = -1.151 \: K/s}}$