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 work done by gas is

Options

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

Answer

B

Explanation

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 = nRT or PDeltaV=nRDeltaT`..(i)
But `PDeltaV` is the work done by the gas
So, `DeltaW = nRT = (2mol)(R)(4T_(0)-T_(0)) = 6RT_(0)`
[from equation (i) ]
`DeltaV prop DeltaV`, i.e., if `DeltaV = 3V_(0), DeltaT = 3T_(0)`.