One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of \(27°C\). If the work done during the process is \(3\, \ kJ\), the final temperature will be equal to \((C_v = 20 JK^{–1})\)
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 150 K.
We know, from the First law of Thermodynamics,
\(\Delta U = q + W\, \ ------(1)\)
We know, for adiabatic condition,
\(q = 0\)
Thus, equation \((1)\) becomes
\(\Delta U = W\, \ -----(2)\)
At constant volume
\(\Delta U = nC_v\Delta T\, \ -----(3)\)
From equation \((1)\) and \((3)\) wehave
\(W = nC_v\Delta T\, \ -------(4)\)
Given,
\(T_i = 27^oC = 27 + 273 = 300\, \ K\)
\(T_f = T\, \ (say)\)
Thus,
\(\Delta T = T_i - T_f\)
or, \(\Delta T = 300 - T\)
\(C_v = 20 JK^{–1}\)
\(W = -3\, \ kJ = -3 \times 10^3\) (∵ Work done by the gas is negative)
Substituting these values in equation \((4)\), we get
\(- 3000 = 1 \times 20 \times (T - 300)\)
\(⇒ T - 300 = -150\)
\(⇒ T = 300 - 150\)
\(⇒ T = 150\, \ K\)