Question

Calculate the maximum work done (in $cal$) when the pressure on 10 grams of hydrogen is reduced from $20$ to $1atm$ at a constant temperature of $273K$ if the gas behaves ideally.

• A. $-8810cal$
• B. $-8180cal$
• C. $+8180cal$
• D. $+8810cal$

Answer 1

The correct option is B $-8180cal$

Maximum work done is obtained when the process is reversible.
So, work done in an isothermal reversible process is given as:
$w=-2.303nRTlog\frac{P_1}{P_2}$

n = number of moles of hydrogen
$n=\frac{\text{given weight}}{\text{molar mass}}$ = $\frac{10}{2}=5mol$

Given : $P_1=20atm$
$P_2=1atm$
$T=273K$
$R=2cal/K.mol$
So,
$w=-2.303\times 5\times 2\times 273\times log\frac{20}{1}$
= $-8179.8Cal$ which is about $-8180Cal$