Ok I got a question
I'm confused by the compression ratio bit in this problem and don't know how to do it
If 3.00 m^3 of a gas initally at STP is placed under a pressure of 3.20 atm, the temperature of the gas rises to 38.0 degrees C. What is the volume?
I did this
p1V1/T1 = p2v2/T2
i then solved for T2 and got about 140 degrees K I think I did it wrong because of the compression ratio how do I do this problem THANKS!!!
Your equation is right but you want to solve for V2.
You should not be solving for T2. They told you what is is already. It is 38 C (315 K)
1 atm*3.00 m^2/273 K = 3.2 atm*V2/315 K
V = (315/273)*(1/3.2)*3.00 = 1.08 m^3
To solve this problem, you can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
In the given problem, the gas is initially at STP (Standard Temperature and Pressure) where the temperature is 273 Kelvin and the pressure is 1 atmosphere. We are given the initial volume, V1 = 3.00 m^3.
We need to find the final volume, V2, when the gas is under a pressure of 3.20 atmospheres and has a temperature of 38.0 degrees Celsius (which we'll convert to Kelvin).
Before proceeding, it's important to note that the compression ratio is not relevant in this problem. It seems to be a misunderstanding or confusion regarding the problem statement.
Now, following the steps to solve the problem:
1. Convert the given temperature from Celsius to Kelvin. Add 273 to the temperature:
T2 = 38.0°C + 273 = 311 Kelvin.
2. Using the ideal gas law, we can set up the following equation:
P1V1 / T1 = P2V2 / T2.
Substituting the known values:
(1 atm) * (3.00 m^3) / (273 K) = (3.20 atm) * V2 / (311 K).
3. Solve for V2 by rearranging the equation:
V2 = (1 atm) * (3.00 m^3) * (311 K) / (3.20 atm) ≈ 93.77 m^3.
Therefore, the volume of the gas at the new conditions is approximately 93.77 m^3.
Remember to double-check your calculations, but this should be the correct approach to solve the problem.