Question

21
178. Calculate the work involved when 1 mol of an ideal gas is compressed
reversibly from 1.00 bar to 5.00 bar at a constant temperature of
300 K.​

Answer 1

Number of moles (n) = 1

P₁ = 1 bar

P₂ = 5 bar

Temperature (T) = 300 K

We know that :-

$\boxed{ \rm \large \: W.D = - 2.303 \times n \times R \times T \times log \dfrac{P_1}{P_2} }$

or

$\boxed{ \rm \large \: W.D = 2.303 \times n \times R \times T \times log \dfrac{P_2}{P_1} }$

Where :- W.D = work

n = number of moles

R = Gas constant

T = temperature

P₁ and P₂ = pressure

$\rm \longrightarrow\large \: W.D = 2.303 \times 1 \times 8.314 \times 300\times log \dfrac{5}{1} \\ \\ \\ \rm \longrightarrow\large \: W.D = 2.303 \times 1 \times 8.314 \times 300\times log 5\\ \\ \\ \rm \longrightarrow\large \: W.D = 2.303 \times 8.314 \times 300\times 0.6989 \\\\ \\ \rm \longrightarrow\large \: W.D = 4014.5 \: J$

Answer :

Work involved when 1 mol of an ideal gas is compressed reversibly from 1.00 bar to 5.00 bar at a constant temperature of 300 K = 4014.5 J

Answer 2

Question :-

→ Calculate the work involved when 1 mol of an ideal gas is compressed

reversibly from 1.00 bar to 5.00 bar at a constant temperature of

300 K.​

Given :-

→ No. of moles ( n ) = 1

→ P₁ = 1.00 bar

→ P₂ = 5.00 bar

→ Temperature ( T ) = 300 K

To Find :-

→ The work done / involved

Solution :-

→ As we know that ,

$\sf -P \Delta V = -2.303 \; nRT \; log \; \dfrac{P_{1}}{P_{2}} \\\\\\\longrightarrow -2.303 \times 1mol \times 8.314 JK^{-1} mol^{-1} \times\;300K log \dfrac{1}{5} \\\\\\\longrightarrow - ( 5744.14 \times (log1-log5) \\\\\longrightarrow (-5774.14 \times ( 0 - 0.699 ) )\\\\\\ \longrightarrow 4015. 1 J \; (answer)$