Question

An ideal gas flows into one end of a porous cylinder at a temperature of 400 K degrees and out the other end at a temperature of 600 K degrees. The input pressure is 20 atmospheres and the output pressure is 10 atmospheres. If the input volume flow rate is 100 cm^3/sec what is the output flow rate?

Answer 1

Answer on Question#40870 – Physics – Molecular Physics | Thermodynamics

An ideal gas flows into one end of a porous cylinder at a temperature of $400\,\mathrm{K}$ degrees and out the other end at a temperature of $600\,\mathrm{K}$ degrees. The input pressure is 20 atmospheres and the output pressure is 10 atmospheres. If the input volume flow rate is $100\,\mathrm{cm}^3/\mathrm{sec}$ what is the output flow rate?

Solution:

$\mathrm{T}_{1} = 400\,\mathrm{K}$ – initial temperature of a gas

$\mathrm{T}_{2} = 600\,\mathrm{K}$ – final temperature of a gas

$\mathrm{p}_{1} = 20\,\mathrm{atm}$ – input pressure of a gas

$\mathrm{p}_{2} = 10\,\mathrm{atm}$ – output pressure of a gas

$\mathrm{V}_{1} = 100\,\mathrm{cm}^{3}$ – input volume of a gas per 1 second

$\mathrm{V}_{2}$ – input volume of a gas per 1 second

We can use the combined gas equation (for a portion of gas per one second)

$$\frac{\mathrm{p}_{1} \mathrm{V}_{1}}{\mathrm{T}_{1}} = \frac{\mathrm{p}_{2} \mathrm{V}_{2}}{\mathrm{T}_{2}}$$ $$\mathrm{V}_{2} = \frac{\mathrm{p}_{1} \mathrm{V}_{1} \mathrm{T}_{2}}{\mathrm{T}_{1} \mathrm{p}_{2}} = \frac{20\,\mathrm{atm} \cdot 100\,\mathrm{cm}^{3} \cdot 600\,\mathrm{K}}{400\,\mathrm{K} \cdot 10\,\mathrm{atm}} = 300\,\mathrm{cm}^{3}$$

Hence, input volume flow rate is $\frac{300\,\mathrm{cm}^{3}}{s} = 300\,\frac{\mathrm{cm}^{3}}{s}$.

Answer: input volume flow rate of the gas is $300\,\frac{\mathrm{cm}^{3}}{s}$.