Question

if 3rd virial coeff is 0.25 L^2 mol^-2, then find volume occupied by 1 mole of real gas at ntp

Answer 1

24.07 L

Answer 2

The virial equation of state for a real gas, for 1 mole of gas ($V_m = V$), can be written as:

$PV = RT \left(1 + \frac{B}{V} + \frac{C}{V^2} + \dots \right)$

where $B$ is the second virial coefficient and $C$ is the third virial coefficient.

Given: Third virial coefficient, $C = 0.25 \, \text{L}^2 \, \text{mol}^{-2}$. The problem provides only the third virial coefficient. In such cases, it is typically assumed that the second virial coefficient ($B$) is zero and higher-order terms are negligible for the purpose of solving the problem. So, the equation simplifies to:

$PV = RT \left(1 + \frac{C}{V^2}\right)$

$PV = RT + \frac{RTC}{V^2}$

We need to find the volume ($V$) occupied by 1 mole of real gas at NTP. NTP (Normal Temperature and Pressure) is generally defined as: Temperature ($T$) = $20^\circ \text{C} = 20 + 273.15 = 293.15 \, \text{K}$ Pressure ($P$) = $1 \, \text{atm}$ Gas constant ($R$) = $0.08206 \, \text{L atm mol}^{-1} \, \text{K}^{-1}$

First, calculate the value of $RT$:

$RT = 0.08206 \, \text{L atm mol}^{-1} \, \text{K}^{-1} \times 293.15 \, \text{K} = 24.06009 \, \text{L atm mol}^{-1}$

Now, substitute the values into the simplified virial equation:

$1 \times V = 24.06009 + \frac{24.06009 \times 0.25}{V^2}$

$V = 24.06009 + \frac{6.0150225}{V^2}$

This can be rearranged into a cubic equation:

$V^3 = 24.06009 V^2 + 6.0150225$

$V^3 - 24.06009 V^2 - 6.0150225 = 0$

To solve this cubic equation, we can use an approximation method, as the correction term for real gases is usually small. As a first approximation, we can assume the gas behaves ideally to estimate $V$. For an ideal gas at NTP:

$V_{ideal} = \frac{RT}{P} = \frac{24.06009 \, \text{L atm mol}^{-1}}{1 \, \text{atm}} = 24.06009 \, \text{L}$

Now, substitute $V \approx V_{ideal}$ into the correction term $\frac{RTC}{V^2}$:

$V \approx V_{ideal} + \frac{RTC}{P V_{ideal}^2}$

$V \approx 24.06009 \, \text{L} + \frac{24.06009 \, \text{L atm mol}^{-1} \times 0.25 \, \text{L}^2 \, \text{mol}^{-2}}{1 \, \text{atm} \times (24.06009 \, \text{L})^2}$

$V \approx 24.06009 \, \text{L} + \frac{6.0150225 \, \text{L}^3 \, \text{atm mol}^{-3}}{578.8836 \, \text{L}^2 \, \text{atm mol}^{-2}}$

$V \approx 24.06009 \, \text{L} + 0.0103907 \, \text{L}$

$V \approx 24.07048 \, \text{L}$

Rounding to a reasonable number of significant figures (e.g., two decimal places, consistent with the precision of R and T):

$V \approx 24.07 \, \text{L}$