Question

1. a rigid cylinder with a movable piston contains a sample of hydrogen gas. at 330. k this sample has a pressure of 150. kpa and a volume of 3.50 l. what is the volume of this sample at stp? look at table a. a) 0.233 l b) 1.96 l c) 4.29 l d) 6.26 l

Explanation:

Step1: Recall the combined - gas law formula

The combined - gas law is $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. At STP, $P_2 = 101.3\ kPa$ and $T_2=273\ K$. Given $P_1 = 150\ kPa$, $V_1 = 3.50\ L$, and $T_1 = 330\ K$.

Step2: Rearrange the combined - gas law formula to solve for $V_2$

$V_2=\frac{P_1V_1T_2}{T_1P_2}$.

Step3: Substitute the given values into the formula

$V_2=\frac{150\ kPa\times3.50\ L\times273\ K}{330\ K\times101.3\ kPa}$.
First, calculate the numerator: $150\times3.50\times273 = 150\times955.5=143325$.
Then, calculate the denominator: $330\times101.3 = 33429$.
$V_2=\frac{143325}{33429}\approx4.29\ L$.

Answer:

C. 4.29 L