A sample containing 2.5\text{ moles} of ammonia gas (\text{NH}_3) is held in a vessel at 4… — JAMB UTME

A sample containing $2.5\text{ moles}$ of ammonia gas ( $\text{NH}_3$ ) is held in a vessel at $400\text{ K}$ under a pressure of $50.0\text{ atm}$ . Under these conditions, the compressibility factor ( $Z$ ) of ammonia is $0.880$ . What is the actual volume occupied by the gas sample in $\text{dm}^3$ ? (Take $R = 0.0821\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ )

Cevap: 1.44 dm^3

Cevap

The actual volume occupied by the ammonia gas sample under the given conditions is

1.44\text{ dm}^3

The compressibility factor

Z

is defined as

Z = \frac{P V}{n R T} = \frac{V_{\text{real}}}{V_{\text{ideal}}}

. Substituting

n = 2.5\text{ mol}

T = 400\text{ K}

P = 50.0\text{ atm}

R = 0.0821\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}

, and

Z = 0.880

into

V = \frac{Z \cdot n R T}{P}

gives

1.44\text{ dm}^3

Adım Adım Çözüm

Identify the relationship between real volume and compressibility factor Z

Z = \frac{P V_{\text{actual}}}{n R T}

The compressibility factor quantifies the deviation of a real gas from ideal gas behavior.

Rearrange the compressibility formula to solve for actual volume (

V_{\text{actual}}

)

V_{\text{actual}} = \frac{Z \cdot n R T}{P}

Isolating

V_{\text{actual}}

allows direct calculation using the provided parameters.

Substitute the given values into the equation and compute the result

V_{\text{actual}} = \frac{0.880 \times 2.5 \times 0.0821 \times 400}{50.0} = 1.44496\text{ dm}^3

Performing the algebraic calculation yields the volume occupied by the real gas.

Anahtar Kavram

Compressibility Factor and Real Gas Deviation

Soru

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