During an adiabatic process, the temperature of 6.50 moles of a monatomic ideal gas drops from 525 °C to 169 °C. What is the work done?

Here is my work, is it correct?

T1 = 525+273 = 798 K

T2 = 169 + 273 = 442 K

work done = R[T2-T1]/1-gamma = 8.314[525 - 798]/(1-1.67) = - 4417.58 J

To calculate the work done during an adiabatic process, you can use the formula:

work done = R * (T2 - T1) / (1 - γ)

where:
- R is the gas constant (8.314 J/(mol·K))
- T2 is the final temperature in Kelvin
- T1 is the initial temperature in Kelvin
- γ (gamma) is the heat capacity ratio (also known as adiabatic index)

In your calculation, you correctly converted the temperatures to Kelvin, which is necessary for the formula. However, there seems to be a small mistake in the value of γ.

For a monatomic ideal gas, the heat capacity ratio γ is equal to 5/3, not 1.67. So, using the correct value:

γ = 5/3 = 1.67

Now, let's calculate the work done:

T1 = 525 °C + 273 = 798 K
T2 = 169 °C + 273 = 442 K
γ = 5/3 = 1.67
R = 8.314 J/(mol·K)

work done = R * (T2 - T1) / (1 - γ)
work done = 8.314 * (442 - 798) / (1 - 1.67)
work done = 8.314 * (-356) / (-0.67)
work done ≈ 4417.58 J

Your final answer of approximately -4417.58 J seems correct, considering that work done during compression is typically negative.