 Thermometry
Question

# Half a mole of helium at ${27}^{0}\mathrm{C}$ and at a pressure of 2 atmosphere is mixed with 1.5 mole of ${\mathrm{N}}_{2}$ at ${77}^{0}$ C and at a pressure of 5 atmosphere so that the volume of the mixture is equal to the sum of their initial volumes. If the temperature of the mixture is ${69}^{0}$C, its pressure in atmosphere is

Moderate
Solution

## PV = nRT  $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$From conservation of mole number${\mathrm{n}}_{1}+{\mathrm{n}}_{2}=\mathrm{n}$${\mathrm{n}}_{1}+{\mathrm{n}}_{2}=\frac{\mathrm{P}}{\mathrm{RT}}\left({\mathrm{V}}_{1}+{\mathrm{V}}_{2}\right)$${\text{n}}_{1}+{\mathrm{n}}_{2}=\frac{\mathrm{P}}{\mathrm{RT}}\left(\frac{{\mathrm{n}}_{1}{\mathrm{RT}}_{1}}{{\mathrm{P}}_{1}}+\frac{{\mathrm{n}}_{2}{\mathrm{RT}}_{2}}{{\mathrm{P}}_{2}}\right)$${\text{n}}_{1}+{\mathrm{n}}_{2}=\frac{\mathrm{P}}{\mathrm{T}}\left(\frac{{\mathrm{n}}_{1}{\mathrm{T}}_{1}}{{\mathrm{P}}_{1}}+\frac{{\mathrm{n}}_{2}{\mathrm{T}}_{2}}{{\mathrm{P}}_{2}}\right)$$\text{0.5+1.5=2=}\frac{\mathrm{P}}{\mathrm{T}}\left(\frac{0.5{\mathrm{T}}_{1}}{{\mathrm{P}}_{1}}+\frac{1.5{\mathrm{T}}_{2}}{{\mathrm{P}}_{2}}\right)$$\text{2=}\frac{\mathrm{P}}{342}\left(\frac{0.5\left(300\right)}{2}+\frac{1.5\left(350\right)}{5}\right)$P = 3.8 atm

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