Question

a sample of a gas has a pressure of 1.26 bar at 58.0°c. what is the pressure of the gas at 118°c?
select the correct answer below:
1.49 bar
1.07 bar
0.672 bar
1.68 bar

Explanation:

Step1: Convert temperatures to Kelvin

$$T_1 = 58.0 + 273.15=331.15\ K$$
$$T_2 = 118 + 273.15 = 391.15\ K$$

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

We know $P_1 = 1.26$ bar, $T_1 = 331.15\ K$, $T_2=391.15\ K$. Rearranging for $P_2$ gives $P_2=\frac{P_1T_2}{T_1}$.

Step3: Calculate $P_2$

$$P_2=\frac{1.26\times391.15}{331.15}$$
$$P_2=\frac{492.849}{331.15}\approx1.49\ bar$$

Answer:

1.49 bar