A 1.80 mol sample of an ideal diatomic gas at a pressure of 1.20 atm and temperature of 410 K undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 710 K and 1.83 atm. (Note: Volume does not stay constant). What is the work done by the gas?
To find the work done by the gas, we can use the formula:
Work = -nRT * ln(V2/V1)
Let's break down the formula and find the values we need.
1. Calculate the change in volume (V2/V1):
From the problem statement, it is mentioned that the volume does not stay constant. We are not given the initial and final volumes, but we know that the pressure and temperature are changing linearly. Since the temperature is directly proportional to the pressure, we can assume that the volume is also directly proportional to the temperature. So, we can write:
(V2/V1) = (T2/T1)
Calculating the ratio of temperatures:
(T2/T1) = (710 K)/(410 K) = 1.7317
2. Calculate the gas constant (R):
The gas constant (R) is a constant value that depends on the units being used. In this case, we are given the pressure in atm, so we need to use the gas constant with the appropriate units, which is 0.0821 L*atm/(mol*K).
3. Calculate the number of moles of gas (n):
We are given that the sample contains 1.80 mol of the gas.
4. Plug the values into the formula and solve for work:
Using the given values:
Work = -nRT * ln(V2/V1)
= -(1.80 mol)(0.0821 L*atm/(mol*K))(410 K) * ln(1.7317)
≈ -360.474 L*atm
Therefore, the work done by the gas is approximately -360.474 L*atm. The negative sign indicates that work is done on the gas during this process.