Question

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

• A. $8192.6cal$
• B. $9258.3cal$
• C. $5924.2cal$
• D. $2896.4cal$

Answer 1

The correct option is A $8192.6cal$

work done in 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,
$w=-2.303\times 5\times 2\times 273\times log\frac{20}{1}$
= $-8192.6Cal$