Question

An ideal gas expands against a constant, external preasure at $2.0$ atmosphere from $20$ litre to $40$ litre and absorbs 10kJ of the energy from surrounding what is the change in internal energy of the system?
(given:$1atm-litre=101.3J)$

• A. $4052J$
• B. $5948J$
• C. $14052J$
• D. $9940J$

Answer 1

Answer: (B)

$5948J$

Solution:- (B) $5948J$

As the heat is absorbed from the surrounding.

Therefore $q=10kJ=10000J$

As we know that,

$w=-P_{ext.}\Delta V$

Given:-

$P_{ext.}=2atm$

$V_i=20L$

$V_f=40L$

$\therefore w=-2(40-20)=-40L-atm=-4052J$

Now from first law of thermodynamics,

$\Delta U=q+w$

$\Rightarrow \Delta U=10000+(-4052)=5948J$

Hence the change in internal energy of the system is $5948J$.