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Chapter 6: Problem 24

A 10.0 g sample of a gas has a volume of \(5.25 \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(762 \mathrm{mm} \mathrm{Hg} .\) If \(2.5 \mathrm{g}\) of the same gas is added to this constant 5.25 L volume and the temperature raised to \(62^{\circ} \mathrm{C},\) what is the new gas pressure?

Short Answer

Expert verified
After solving the equations, the new gas pressure will be the value of \(P_2\) as found in Step 3.

Step by step solution

01

Finding the Number of Moles for Initial Conditions

Use the ideal gas law to calculate the initial number of moles. First, convert the temperature into Kelvin by adding 273.15 to the degrees Celsius. The formula was \(P_1V_1 = n_1RT\) where \(P_1 = 762 \mathrm{mm} \mathrm{Hg}\), \(V_1 = 5.25 \mathrm{L}\), \(R = 62.364 \mathrm{L.mmHg/mol.K}\) and \(T_1 = 25^{\circ} \mathrm{C} = 298.15 \mathrm{K}\). Solving for \(n_1\), we get \(n_1 = P1*V1/(R*T1)\)
02

Adding More Gas

2.5g of the gas are added to the sample. The molar mass of the gas is given by relation between mass and moles from step 1. Calculate the number of moles of the added gas (\(n_2 = m/M\)), where m is the mass of added gas and M is the molar mass from step 1. The total number of moles (\(n_{total}\)) is the sum of the initial and the added moles, \(n_{total} = n_1 + n_2\)
03

Finding the New Gas Pressure

The new pressure (\(P_2\)) is found by solving the ideal gas law again with the new temperature, volume and number of moles. We convert the new temperature to Kelvin, \(T_2 = 62^{\circ} \mathrm{C} = 335.15 \mathrm{K}\). The volume (\(V_2\)) remains the same at \(5.25 \mathrm{L}\). Then solve for \(P_2\), \(P_2 = n_{total}*R*T_2/V_2\)

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Most popular questions from this chapter

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