Two moles of an ideal gas is contained in a cylinder fitted with a frictionless movalbe piston, exposed to the atmosphere, at an initial temperature `T_(0)`. The gas is slowly heated so that its volume becomes fout times the initial value. The wor done by gas is

Options

A. zero
B. `2 RT_(0)`
C. `4 RT_(0)`
D. `6 RT_(0)`

Answer

D

Explanation

d. Since the gas is slowly heated, it remains in equilibrium (more or less) with the atmosphere, i.e., the process takes place at a constant pressure.
Now, from `PV = nRT, PdV = nRdT`
or, `P Delta V = nR Delta T`
But `P Delta V` is the work done by the gas.
So, `Delta W = nR Delta T = (2 mol) (R ) (4 T_(0) - T_(0)) = 6 RT_(0)`
[From Eq. (i) `Delta V prop Delta T`, i.e., if `Delta V = 3V_(0), Delta T = 3 T_(0)`]