A gas has a compressibility factor of 0.5 and a molar volume of 0.4 dm³ mol⁻¹ at a temperature of 800 K and pressure $x$ atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be $\gamma$ dm³ mol⁻¹. The value of $\frac{x}{\gamma }$ is:

Correct Answer: The value of $\frac{x}{\gamma }$ is 100.

Step-by-Step Explanation:

Step 1: Determine the ideal molar volume ($\gamma$)

The compressibility factor ($Z$) of a real gas is defined as the ratio of its actual molar volume ($V_{\mathrm{m,}\text{real}}$) to its ideal molar volume ($V_{\mathrm{m,}\text{ideal}}$) at the same temperature and pressure conditions:

$Z=\frac{V_{\mathrm{m,}\text{real}}}{V_{\mathrm{m,}\text{ideal}}}$

Given that $Z=0.5$, $V_{\mathrm{m,}\text{real}}=0.4{\text{dm}}^3{\text{mol}}^{-1}$, and $V_{\mathrm{m,}\text{ideal}}=\gamma {\text{dm}}^3{\text{mol}}^{-1}$:

$0.5=\frac{0.4}{\gamma }$

$\gamma =\frac{0.4}{0.5}=0.8{\text{dm}}^3{\text{mol}}^{-1}$

Step 2: Calculate the pressure ($x$)

Using the ideal gas equation for one mole of gas:

$P·V_{\mathrm{m,}\text{ideal}}=RT$

Substituting $P=x\text{atm}$, $V_{\mathrm{m,}\text{ideal}}=0.8{\text{dm}}^3{\text{mol}}^{-1}$, $T=800\text{K}$, and using $R=0.08{\text{dm}}^3\text{atm K}{\mathrm{-1}}^{}{\text{mol}}^{-1}$:

$x·0.8=0.08\times 800$

$x·0.8=64$

$x=\frac{64}{0.8}=80\text{atm}$

Step 3: Find the value of $\frac{x}{\gamma }$

Now, calculate the ratio of pressure to ideal molar volume:

$\frac{x}{\gamma }=\frac{80}{0.8}=100$

Thus, the value of $\frac{x}{\gamma }$ is 100.