Question

a gas in a rigid container has a pressure of 2.0 atm at a temperature of 25°c. if the temperature is increased to 100°c while the volume remains constant, what will be the new pressure of the gas? the new pressure would be select one select one

Explanation:

Step1: Convert temperatures to Kelvin

For initial temperature: $T_1 = 25 + 273 = 298\ K$
For final temperature: $T_2 = 100 + 273 = 373\ K$

Step2: Apply Gay - Lussac's Law ($\frac{P_1}{T_1}=\frac{P_2}{T_2}$)

We know $P_1 = 2.0\ atm$, $T_1 = 298\ K$, $T_2 = 373\ K$. Rearranging for $P_2$ gives $P_2=\frac{P_1\times T_2}{T_1}$
Substitute values: $P_2=\frac{2.0\times373}{298}$
Calculate: $P_2=\frac{746}{298}\approx2.5\ atm$

Answer:

$2.5\ atm$