Question

One mole of a non - ideal gas undergoes a change of state (2.0 atm, 3.0L, 95 K ) $\to$ (4.0 atm, 5.0 L, 245 K) with a change in internal energy , $\Delta U=30L-atm$. The change in enthalpy of the process in L-atm is:

• A. 40
• B. 42.3
• C. 44
• D. none of these

Answer 1

The correct option is C 44

Solution:- (C) 44.0

Given:-

$(2\text{atm},3.0\text{L},95\text{K})\to (4\text{atm},5.0\text{L},245\text{K})$

Therefore,

$\begin{array}{cc}{P}_1=2atm& P_2=4atm\\ V_1=3L& V_2=5L\\ T_1=95K& T_2=245K\end{array}$

$\Delta U=30L-atm$

As we know that,

$\Delta H=\Delta U+(P_2{V}_2-P_1{V}_1)$

$\therefore \Delta H=30+(4\times 5-2\times 3)$

$\Rightarrow \Delta H=30+14=44L-atm$

Hence the change in enthalpy is $44L-atm$.