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Chapter 6: Problem 105

Gas cylinder A has a volume of 48.2 L and contains \(\mathrm{N}_{2}(\mathrm{g})\) at 8.35 atm at \(25^{\circ} \mathrm{C} .\) Gas cylinder \(\mathrm{B},\) of unknown volume, contains \(\mathrm{He}(\mathrm{g})\) at 9.50 atm and \(25^{\circ} \mathrm{C} .\) When the two cylinders are connected and the gases mixed, the pressure in each cylinder becomes 8.71 atm. What is the volume of cylinder \(\mathrm{B} ?\)

Short Answer

Expert verified
To find the volume of cylinder B, follow the steps outlined above, substituting the given values into the appropriate formulas. The final answer will depend on these specific values.

Step by step solution

01

Calculate the number of moles of gas in cylinder A

From the given values, apply the ideal gas law to calculate the number of moles of gas, \(n_A\), in cylinder A. \(P_A = 8.35 \,atm\), \(V_A = 48.2 \,L\), \(R = 0.0821 \, atm \cdot L/(mol \cdot K)\), and \(T_A = 25°C + 273.15 = 298.15 \,K\). Therefore, \(n_A = \frac{P_AV_A}{RT_A}\).
02

Calculate the total final moles of the mixed gases

We know that after mixing the gases from both cylinders, the final pressure, \(P_f\), is 8.71 atm. So the total moles of gases, \(n_{total}\), is given by \(n_{total} = n_A + n_B = \frac{P_f V_total}{R T_f}\).
03

Calculate the volume of cylinder B

We know that \(V_total = V_A + V_B\). Substituting this into the equation from Step 2 and simplifying, we find that \(n_B = \frac{P_fV_B}{R T_f}\). Subtracting \(n_A\) from both sides of the equation gives \(n_B\). Now, we can find the volume of cylinder B by rearranging the formula: \(V_B = \frac{(n_{total} - n_A)RT_f}{P_f}\).

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Most popular questions from this chapter

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