Question

A sealed flask with a capacity of $2\, dm ^3$ contains $11 \, g$ of propane gas The flask is so weak that it will burst if the pressure becomes $2\, MPa$ The minimum temperature at which the flask will burst is ______${ }^{\circ} C$ [Nearest integer] 

(Given: $R =8.3 \,J \,K ^{-1} mol ^{-1}$ Atomic masses of $C$ and $H$ are $12\, u$ and $1 \,u$ respectively) (Assume that propane behaves as an ideal gas)

Answer 1

Correct Answer: 1655

To determine the minimum temperature at which the flask will burst, we use the ideal gas law: PV = nRT. Here, P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. We need to find T when P = 2 MPa, V = 2 dm³, and the mass of propane is 11 g.

First, calculate the molar mass of propane (C₃H₈):

Molar mass = 3(12 u) + 8(1 u) = 44 u (g/mol)

Next, calculate the number of moles, n, of propane:

n = mass / molar mass = 11 g / 44 g/mol = 0.25 mol

Convert the volume from dm³ to m³: 2 dm³ = 0.002 m³

Convert pressure from MPa to Pa: 2 MPa = 2 × 10⁶ Pa

Plug these values into the ideal gas law and solve for T:

T = PV / nR = (2 × 10⁶ Pa) × (0.002 m³) / (0.25 mol × 8.3 J K^-1 mol^-1)

T = 192771.08 K

Convert T from Kelvin to Celsius:

T_°C = 192771.08 K - 273.15 ≈ 1655°C

Thus, the minimum temperature at which the flask will burst is 1655°C. This value fits the provided range of 1655,1655.