To find the new volume of the gas sample, we can use the combined gas law equation, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law into one formula. The combined gas law equation is as follows:
\( \frac{{P_1 \cdot V_1 \cdot T_2}}{{n_1}} = \frac{{P_2 \cdot V_2 \cdot T_1}}{{n_2}} \)
Given values:
- \( P_1 = 1.5 \, \text{atm} \)
- \( V_1 = 1.0 \, \text{L} \)
- \( T_1 = 25^\circ C = 25 + 273 = 298 \, \text{K} \)
- \( n_1 = 2.3 \, \text{moles} \)
First, we need to calculate the initial volume occupied by 2.3 moles of gas at 1.5 atm and 298 K. Then, we can find the final volume when 1.1 moles are added, and the pressure and temperature change.
1. Calculate the initial volume using the given initial conditions:
\( \frac{{1.5 \cdot 1.0 \cdot 298}}{{2.3}} = V_1 \)
\( V_1 \approx 184.6 \, \text{L} \)
2. Next, determine the final volume when 1.1 moles are added, and the pressure and temperature change:
- \( P_2 = 6.0 \, \text{atm} \)
- \( T_2 = 100^\circ C = 100 + 273 = 373 \, \text{K} \)
- \( n_2 = 2.3 + 1.1 = 3.4 \, \text{moles} \)
Substitute the known values into the combined gas law equation and solve for \( V_2 \):
\( \frac{{1.5 \cdot 1.0 \cdot 373}}{{3.4}} = 6.0 \cdot V_2 \cdot 298 \)
\( 559.5 = 1788 \cdot V_2 \)
\( V_2 = \frac{{559.5}}{{1788}} \)
\( V_2 \approx 0.31 \, \text{L} \)
Therefore, the new volume of the gas sample would be approximately 0.31 liters when 1.1 moles are added, and the pressure is changed to 6.0 atm and the temperature to 100°C.
