Question

A balloon that can hold 85 L of air is inflated with 7.056 grams of H2 gas at a pressure of 101.3 kPa. What is the temperature, in Celsius, of the balloon.

Answer 1

To find the temperature in Celsius of a balloon inflated with 7.056 grams of H2 gas, we use the Ideal Gas Law, calculate the moles of gas, and rearrange the formula to solve for temperature. After calculation, the temperature is determined to be 17.7°C.

Step-by-step explanation:

The student asks for the temperature in Celsius of a balloon that is inflated with 7.056 grams of H2 gas and holds 85 L of air at a pressure of 101.3 kPa. To find the temperature, we can use the Ideal Gas Law formula, which is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, calculate the number of moles (n) of H2 gas using its molar mass (2.016 g/mol). The calculation is n = 7.056 g / 2.016 g/mol = 3.5 moles. Next, rearrange the Ideal Gas Law to solve for T (temperature in Kelvin): T = PV / (nR). Substituting the values into the equation, we get T = (101.3 kPa × 85 L) / (3.5 mol × 8.314 L·kPa/mol·K) = 290.85 K. Finally, convert the temperature from Kelvin to Celsius by subtracting 273.15 from the Kelvin temperature: 290.85 K - 273.15 = 17.7°C.

Answer 2

Temperature is 258.32°C

Step-by-step explanation:

Using the ideal gas equation;

PV = nRT --------------(i)

Where;

P = Pressure of the gas

V = Volume of the gas

n = number of moles of the gas

R = Gas constant = 8.31 J/mol · K

T = Temperature

Given:

mass of H₂ gas = 7.056 grams

Volume of the gas = 85L = 8.5 x 10⁻³m³

Pressure of the gas = 101.3kPa = 101.3 x 10³Pa = 1.013 x 10⁵Pa

Steps:

(i) Using the mass of the gas, calculate the number of moles using the relation:

n = m / M ----------------- (ii)

Where;

m = mass of H₂ = 7.056g

M = Molar mass of H₂ = 1g/mol

Substitute these values into equation (ii) as follows:

n = 7.056g / (1g/mol)

n = 7.056mol

(ii) Now calculate the temperature of the balloon by substituting the necessary values into equation (i)

(1.013 x 10⁵Pa)(8.5 x 10⁻³m³) = (7.056mol) (8.31 J/mol · K)(T)

T = (1.013 x 10⁵Pa)(8.5 x 10⁻³m³) ÷ (7.056mol) (8.31 J/mol · K)

Solving the above gives

T = 14.68K

Convert this to Celsius

T = 273 - 14.68

T = 258.32°C