Question

If 3.00 m^3 of a gas initially at STP is placed under a pressure of 4.00 atm, and the temperature of the gas rises to 38.0°C, what is the volume?

a) 3.00 m^3
b) 4.00 m^3
c) 2.25 m^3
d) 2.50 m^3

Answer 1

Using the combined gas law and converting temperatures to Kelvin, the final volume of a gas initially at 3.00 m³ under STP conditions then placed under 4.00 atm and heated to 38.0°C is calculated to be approximately 2.25 m³.

Step-by-step explanation:

If 3.00 m³ of a gas initially at STP (Standard Temperature and Pressure) is placed under a pressure of 4.00 atm, and the temperature of the gas rises to 38.0°C, to find the final volume of the gas we can use the combined gas law which is:

P1V1/T1 = P2V2/T2

Where P1 and V1 are the initial pressure and volume, T1 is the initial temperature, and P2, V2, and T2 are the final pressure, volume, and temperature respectively. Remember to convert temperatures to Kelvin.

First, convert the initial temperature of STP (0°C) to Kelvin by adding 273.15, which gives us 273.15 K. Now, convert the final temperature of 38.0°C to Kelvin by the same method, giving us 311.15 K.

Now we can apply the combined gas law:

(1.00 atm)(3.00 m³) / (273.15 K) = (4.00 atm)(V2) / (311.15 K)

Rearrange to solve for V2:

V2 = (3.00 m³)(4.00 atm)(273.15 K) / (1.00 atm)(311.15 K)

V2 = (3.00 m³)(4.00)(273.15) / (311.15)

V2 ≈ 2.25 m³

Hence, the correct answer is (c) 2.25 m³.