Question

the temperature in a 200 ml constant volume steel container filled with argon gas is 288.15 k and the pressure is 98.0 kpa. if the volume remains the same, what would be the pressure if the temperature were increased to 338.15 k?
a 425 kpa
b 115 kpa
c 993 kpa
d 83.5 kpa

Explanation:

Step1: Use Gay - Lussac's Law

Gay - Lussac's Law is \( \frac{P_1}{T_1}=\frac{P_2}{T_2}\), where \(P_1 = 98.0\ \text{kPa}\), \(T_1=288.15\ \text{K}\), \(T_2 = 338.15\ \text{K}\), and \(P_2\) is the unknown pressure.

Step2: Solve for \(P_2\)

Rearrange the formula to \(P_2=\frac{P_1\times T_2}{T_1}\). Substitute the values: \(P_2=\frac{98.0\times338.15}{288.15}\).
First, calculate \(98.0\times338.15 = 98\times(300 + 38.15)=98\times300+98\times38.15=29400+3738.7 = 33138.7\).
Then, \(P_2=\frac{33138.7}{288.15}\approx115\ \text{kPa}\).

Answer:

B. 115 kPa