One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be :

Question

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be : (A) 300 R (B) 300 R ln 6 (C) 300 R ln 2 (D) 300 R ln 7

Tardigrade Admin · Accepted Answer

Work done in isothermal process is given as

W = - n RT ln \frac{V _{2}}{V _{1}}

.
Using

P_{1} V_{1} = P_{2} V_{2}

,
we get

\frac{V _{2}}{V _{1}} = \frac{P _{1}}{P _{2}}

.
Therefore,

W = - n RT ln \frac{P _{1}}{P _{2}}

Now given that

T = 27^{\circ} C = 300 K

(∵ 0^{\circ} C = 273 K \Rightarrow 27^{\circ} C \Rightarrow 27 + 273 K)

and

P_{2} = 2 P_{1}

Therefore,

W = - 300 n R ln \frac{P _{1}}{2 P _{1}}

Also,

n = 1 \Rightarrow W = - 300 R ln (\frac{1}{2})

.
Now, using

ln (\frac{a}{b}) = ln (a) - ln (b)

, we get

W = - 300 R [ln (1) - ln (2)] = - 300 R [0 - ln 2]

\Rightarrow W = 300 R ln 2

Q. One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27∘C. The work done on the gas will be :

300 R

300 R ln 6

300 R ln 2

300 R ln 7

Q. One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, $2 7^{\circ} C$ . The work done on the gas will be :