Questions: A glass container was initially charged with 2.30 moles of a gas sample at 4.00 atm and 21.7°C. Some of the gas was released as the temperature was increased to 29.1°C, so the final pressure in the container was reduced to 1.05 atm. How many moles of the gas sample are present at the end?

Solution Steps

We are given the initial and final conditions of the gas in the container:

• Initial moles of gas, \( n_1 = 2.30 \) moles
• Initial pressure, \( P_1 = 4.00 \) atm
• Initial temperature, \( T_1 = 21.7^{\circ} \mathrm{C} = 21.7 + 273.15 = 294.85 \) K
• Final pressure, \( P_2 = 1.05 \) atm
• Final temperature, \( T_2 = 29.1^{\circ} \mathrm{C} = 29.1 + 273.15 = 302.25 \) K

The ideal gas law is given by: \[ PV = nRT \] Since the volume \( V \) and the gas constant \( R \) are constant, we can use the combined gas law to relate the initial and final states: \[ \frac{P_1 V}{T_1} = n_1 R \] \[ \frac{P_2 V}{T_2} = n_2 R \] Dividing these two equations, we get: \[ \frac{P_1}{T_1} \cdot \frac{T_2}{P_2} = \frac{n_1}{n_2} \]

Rearrange the equation to solve for \( n_2 \): \[ n_2 = n_1 \cdot \frac{P_2}{P_1} \cdot \frac{T_1}{T_2} \] Substitute the known values: \[ n_2 = 2.30 \cdot \frac{1.05}{4.00} \cdot \frac{294.85}{302.25} \]

Perform the calculation: \[ n_2 = 2.30 \cdot 0.2625 \cdot 0.9755 \] \[ n_2 \approx 0.5880 \]

Final Answer