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2 years ago

What is the pressure in atm exerted by 1.8 g of H_{2} gas exert in a 4.3 L balloon at 27°C? R =; 0.821(L^ * atm)/(mol^ * K)

Chemistry

1 answer:

trapecia [35]2 years ago

6 0

Answer:

5.12 atm

Explanation:

Before you can use the Ideal Gas Law to find the pressure, you need to 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 plug 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

**Based on my past experiences, I believe the constant (R) you provided may have been mistyped. Instead of 0.821, I used 0.0821.**

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