Calculate the work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K

Question

Calculate the work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K (A) -4138.8J (B) -4938.8J (C) -5744.1J (D) -6257.2J

Tardigrade Admin · Accepted Answer

Given that P1=10atm , P2=1atm , T=300K , n=1 R=8.314J/K/mol Now, by using W=2.303nRTlog10(P2/P1) =2.303× 1× 8.314× 300log10(1/10) W= -5744.1Joule

Q. Calculate the work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at $300 K$

$- 4138.8 J$

$- 4938.8 J$

$- 5744.1 J$

$- 6257.2 J$

Given that
$P_{1} = 10 a t m$ ,
$P_{2} = 1 a t m$ ,
$T = 300 K$ ,
$n = 1$
$R = 8.314 J / K / m o l$
Now, by using
$W = 2.303 n RTl o g_{10} \frac{P _{2}}{P _{1}}$
$= 2.303 \times 1 \times 8.314 \times 300 l o g_{10} \frac{1}{10}$
$W = - 5744.1 J o u l e$

Q. Calculate the work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300K

−4138.8J

−4938.8J

−5744.1J

−6257.2J