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What is the pressure (in atmospheres) exerted by a 0.500 mol sample of N2 gas in a 10.0 L container at 298 K?

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To solve this problem, we can use the Ideal Gas Law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas. The Ideal Gas Law is expressed as:

PV = nRT

where R is the gas constant.

R = 0.0821 L·atm/mol·K

We can rearrange this equation to solve for pressure:

P = nRT/V

where:

n = 0.500 mol (number of moles of N2 gas)

R = 0.0821 L·atm/mol·K (gas constant)

T = 298 K (temperature)

V = 10.0 L (volume)

Substituting the values:

P = (0.500 mol)(0.0821 L·atm/mol·K)(298 K)/(10.0 L)

P = 12.1 atm

Therefore, the pressure exerted by a 0.500 mol sample of N2 gas in a 10.0 L container at 298 K is 12.1 atm.

User Mushif Ali Nawaz
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