If 2.42 m³ of a gas initially at STP is placed under a pressure of 3.80 atm, and the temperature of the gas rises to 37.9 °C, what is the new volume of the gas?

To solve this problem, we can apply the combined gas law, which relates pressure, volume, and temperature of a gas. The formula for the combined gas law is:

[tex]\frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2}[/tex]

Where:

[tex]P_1[/tex] = initial pressure
[tex]V_1[/tex] = initial volume
[tex]T_1[/tex] = initial temperature (in Kelvin)
[tex]P_2[/tex] = final pressure
[tex]V_2[/tex] = final volume
[tex]T_2[/tex] = final temperature (in Kelvin)

Given in the problem:

Initial volume [tex]V_1 = 2.42 \text{ m}^3[/tex]
Initial pressure [tex]P_1 = 1 \text{ atm}[/tex] at STP
Initial temperature [tex]T_1 = 273.15 \text{ K}[/tex] at STP (since STP conditions assume a temperature of 0°C or 273.15 K)
Final pressure [tex]P_2 = 3.80 \text{ atm}[/tex]
Final temperature in Celsius: 37.9°C. Convert this to Kelvin by adding 273.15:

[tex]T_2 = 37.9 + 273.15 = 311.05 \text{ K}[/tex]

We need to find the final volume [tex]V_2[/tex]. Rearrange the combined gas law formula to solve for [tex]V_2[/tex]:

[tex]V_2 = \frac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot P_2}[/tex]

Substitute the values into the equation:

[tex]V_2 = \frac{1 \text{ atm} \times 2.42 \text{ m}^3 \times 311.05 \text{ K}}{273.15 \text{ K} \times 3.80 \text{ atm}}[/tex]

Calculate [tex]V_2[/tex]:

[tex]V_2 = \frac{2.42 \times 311.05}{273.15 \times 3.80}[/tex]
[tex]V_2 \approx \frac{753.211}{1037.97}[/tex]
[tex]V_2 \approx 0.726 \text{ m}^3[/tex]

Therefore, under the new conditions of 3.80 atm pressure and a temperature of 37.9°C, the volume of the gas is approximately 0.726 cubic meters.