Questions: A sample of gas in a flexible container initially occupies a volume of 3.65 L at 271 K. Then, the volume of the gas inside the container decreases to 2.99 L at 290 K and 1.76 atm. The number of moles of the gas remains constant.

Solution Steps

The given values are: $V_1 = 3.65 L$ $T_1 = 271 K$ $V_2 = 2.99 L$ $T_2 = 290 K$ $P_2 = 1.76 atm$

We need to find the initial pressure $P_1$.

Since the number of moles of gas remains constant, we can use the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

We rearrange the equation to solve for $P_1$: $P_1 = \frac{P_2V_2T_1}{V_1T_2}$

$P_1 = \frac{1.76 \text{ atm} \times 2.99 \text{ L} \times 271 \text{ K}}{3.65 \text{ L} \times 290 \text{ K}}$ $P_1 = \frac{1433.7584}{1058.5}$ $P_1 = 1.35 \text{ atm}$

Final Answer