**A given amount of gas occupies 10.0dm5 at 4atm and 273°C. The number of moles of...**

### Question

A given amount of gas occupies 10.0dm5 at 4atm and 273°C. The number of moles of the gas present is [Molar volume of gas at s.t.p = 22.4dm\(^3\)]

### Options

Related Lesson: Key Concepts and Summary | Gases

The correct answer is A.

### Explanation:

To solve this question, we'll use the Ideal Gas Law, which is given by the formula:

\[PV = nRT\]

Where *P* is pressure, *V* is volume, *n* is the number of moles, *R* is the gas constant, and *T* is temperature in Kelvin.

First, let's convert the given temperature from Celsius to Kelvin:

\[T = 273 + 273 = 546\,\text{K}\]

Next, we'll use the given values and the Ideal Gas Law to find the number of moles:

\[4\,\text{atm} \times 10\,\text{dm}^3 = n \times 0.0821\,\frac{\text{atm}\cdot\text{dm}^3}{\text{mol}\cdot\text{K}} \times 546\,\text{K}\]

Solve for *n*:

\[n = \frac{4\,\text{atm} \times 10\,\text{dm}^3}{0.0821\,\frac{\text{atm}\cdot\text{dm}^3}{\text{mol}\cdot\text{K}} \times 546\,\text{K}} \approx 0.89\,\text{mol}\]

Therefore, the number of moles of the gas present is approximately 0.89 mol, which corresponds to **Option A**.

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PV = nRT

R = 0.08206(gas constant)

T = 273 + 273 = 546k(kelvin)

Therefore, n = PV/RT

n = 4 x 10/ 546 x 0.08206 = 0.8869, approx. 0.89mol