A sample of gas (1.9 mol)is in a flask at 21 °C and 697 mm Hg.The flask is opened and more gas is added to the flask. The new pressure is 841 mm Hg and the temperature is now 26 °C.There are now ________ mol of gas in the flask.
A)1.6
B)2.3
C)2.9
D)3.5
E)0.28

Question

Quizplus · Accepted Answer

Using the ideal gas law and the given conditions, we can calculate the moles of gas after the addition. Initially, we have $P_1 = 697 \, \text{mm Hg}$, $T_1 = 21 \, °C = 294 \, K$, and $n_1 = 1.9 \, \text{mol}$. After adding more gas, $P_2 = 841 \, \text{mm Hg}$ and $T_2 = 26 \, °C = 299 \, K$. We can use the combined gas law, rearranged to solve for $n_2$: $\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$. Assuming the volume $V$ remains constant and canceling it out, we get $n_2 = \frac{P_2T_1n_1}{P_1T_2}$. Plugging in the values, $n_2 = \frac{841 \times 294 \times 1.9}{697 \times 299}$, which calculates to approximately 2.3 mol.

A Sample of Gas (1