Chemistry - Thermodynamics - I Question with Solution | TestHub

Given:

State-I: $P_1=1\times 10^{5}Pa$, $T_1=300K$

State-II: $P_2=2\times 10^{5}Pa$, $T_2=600K$

$n=1$ mole

$a\approx 0$

$b=50mLmo{l}^{-1}=50\times 10^{-6}{m}^{3}mo{l}^{-1}$

$C_p=20J{K}^{-1}mo{l}^{-1}$

$R=8.314J{K}^{-1}mo{l}^{-1}$ (standard gas constant)

The change in enthalpy $\Delta H$ can be expressed as:

$$\Delta H={\int }_{T_1}^{T_2}{C}_{p}dT+{\int }_{P_1}^{P_2}\left[V-T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP$$

For a van der Waals gas, the equation of state is:

$$\left(P+\frac{a{n}^2}{V^2}\right)(V-nb)=nRT$$

Since $a\approx 0$ and for one mole ($n=1$):

$$P(V-b)=RT$$

$$V-b=\frac{RT}{P}$$

$$V=\frac{RT}{P}+b$$

Now, let's find ${\left(\frac{\partial V}{\partial T}\right)}_P$:

$${\left(\frac{\partial V}{\partial T}\right)}_P=\frac{R}{P}$$

Substitute this into the given partial derivative expression:

$$V-T{\left(\frac{\partial V}{\partial T}\right)}_P=\left(\frac{RT}{P}+b\right)-T\left(\frac{R}{P}\right)=\frac{RT}{P}+b-\frac{RT}{P}=b$$

So, the enthalpy change becomes:

$$\Delta H={\int }_{T_1}^{T_2}{C}_{p}dT+{\int }_{P_1}^{P_2}bdP$$

$$\Delta H=C_p(T_2-T_1)+b(P_2-P_1)$$

Now, plug in the values:

$C_p=20J{K}^{-1}mo{l}^{-1}$

$T_2-T_1=600K-300K=300K$

$b=50\times 10^{-6}{m}^{3}mo{l}^{-1}$

$P_2-P_1=2\times 10^{5}Pa-1\times 10^{5}Pa=1\times 10^{5}Pa$

$$\Delta H=(20J{K}^{-1}mo{l}^{-1})(300K)+(50\times 10^{-6}{m}^{3}mo{l}^{-1})(1\times 10^{5}Pa)$$

$$\Delta H=6000Jmo{l}^{-1}+5Jmo{l}^{-1}$$

$$\Delta H=6005Jmo{l}^{-1}$$

The change in enthalpy for one mole of the gas is $6005J$.

The final answer is $\boxed{6005J}$.