Question

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27c. The work done on the gas will be : (1)300 r (2)300 r ln 6 (3)300 r ln 2 (4)300 r ln 7

Answer 1

we know, workdone on the gas is given by, $w=P\int{dv}$

for isothermal condition, T is constant.
so, P = nRT/V

and $w=\frac{nRT}{V}\int{dv}$

or, $w=nRT\int{\frac{dV}{V}}$

or, $w=nRTln\frac{V_f}{V_i}$

as temperature is constant. $P_iV_i=P_fV_f$

so, formula of workdone = $nRTln\frac{P_i}{P_f}$

given, n = 1 , T = 27°C = 300K and $P_f=2P_i$

so, W = 1 × R × 300ln2 = 300Rln2

hence, option (3) is correct.