Answer :

The pressure in atm exerted by 1.8 g of H₂ gas exert in a 4.3 L balloon at 27ºC is 5.12 atm

What is Ideal gas law ?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Let's convert grams to moles (via molar mass).

Molar Mass (H₂) : 2 (1.008 g/mol)

Molar Mass (H₂) : 2.016 g/mol

1.8 grams H₂ 1 mole

---------------------- x ---------------------- = 0.893 moles H₂

2.016 grams

The Ideal Gas Law equation looks like this:

PV = nRT

In this equation,

  • P = pressure (atm)
  • V = volume (L)
  • n = moles
  • R = Ideal Gas Constant (0.0821 L*atm/mol*K)
  • T = temperature (K)

After converting Celsius to Kelvin, you can put the given values into the equation and simplify to find the pressure.

P = ? atm R = 0.0821 L*atm/mol*K

V = 4.3 L T = 27 °C + 273.15 = 300.15 K

n = 0.893 moles

PV = nRT

P (4.3 L) = (0.893 moles) (0.0821 L*atm/mol*K)( 300.15 K)

P (4.3 L) = 22.0021

P = 5.12 atm

Therefore, The pressure in atm exerted by 1.8 g of H₂ gas exert in a 4.3 L balloon at 27ºC is 5.12 atm

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Final answer:

The pressure exerted by 1.8 g of H2 gas in a 4.3 L balloon at 27°C is calculated to be approximately 0.547 atm using the ideal gas law.

Explanation:

To calculate the pressure exerted by hydrogen gas in a balloon using the ideal gas law, PV = nRT, we first need to determine the number of moles of H2 gas. Since the molecular weight of H2 is 2.01588 g/mol, we can calculate the moles of H2:

n = mass (g) / molecular weight (g/mol) = 1.8 g / 2.01588 g/mol = 0.8928 mol

Now, converting the temperature to Kelvin:

T = 27°C + 273.15 = 300.15 K

Using the given value of R = 0.0821 (L·atm)/(mol·K), we can rearrange the ideal gas law to solve for pressure (P):

P = (nRT) / V

P = (0.8928 mol × 0.0821 (L·atm)/(mol·K) × 300.15 K) / 4.3 L

P = 0.547 atm