A gas consisting of 20.0 moles of monatomic ideal gas has a change in entropy of 200 J/K when taken through a process that starts at with a volume of 6.00 m³ at 300 K and ends with a volume 2.00 m³. The molar specific heat of the gas is 3R/2 for constant volume processes. What is the final temperature of the gas?

Question

Quizplus · Accepted Answer

The change in entropy is given by:$$\Delta S = nC_v\ln(\frac{V_f}{V_i}) + nR\ln(\frac{T_f}{T_i})$$where $n$ is the number of moles, $C_v$ is the molar specific heat at constant volume, $V_i$ and $V_f$ are the initial and final volumes, $T_i$ and $T_f$ are the initial and final temperatures, and $R$ is the ideal gas constant.Substituting the given values, we get:$$200 J/K = 20.0 mol 	imes \frac{3R}{2} \ln(\frac{2.00 m^3}{6.00 m^3}) + 20.0 mol 	imes R\ln(\frac{T_f}{300 K})$$Simplifying, we get:$$200 J/K = 60.0 R \ln(\frac{1}{3}) + 20.0 R\ln(\frac{T_f}{300 K})$$$$200 J/K = -60.0 R(1.0986) + 20.0 R\ln(\frac{T_f}{300 K})$$$$200 J/K = -65.92 R + 20.0 R\ln(\frac{T_f}{300 K})$$$$20.0 R\ln(\frac{T_f}{300 K}) = 265.92 R$$$$\ln(\frac{T_f}{300 K}) = 13.296$$$$\frac{T_f}{300 K} = e^{13.296}$$$$T_f = 300 K 	imes e^{13.296} = 1390 K$$Therefore, the final temperature of the gas is 1390 K.

A Gas Consisting of 20