Question

Баллон емкостью 74,7 л содержит 1,98 кг углекислого газа при температуре 27 ° С. Баллон выдерживает давление не свыше 3000 кПа. При какой температуре он может разорваться? Молярную массу углекислого газа принять равной 0,044 кг/моль.

Answer 1

Дано    V=74,7*10^-3 м3    m=1,98 кг     T1=300 К       P2=3*10^6 Па

P1*V1=m*R*T1/M
P1=m*R*T/V1*M=1,98*8,31*300/74,7*44*10^-3=1, 5  МПа
из уравнения состояния
при V= const
P1/T1=P2/T2
T2=P2*T1/P1=3*300/1,5=600 К
t2=600-273=327C

Answer 2

Calculation of Temperature at which the Cylinder can Burst

To determine the temperature at which the cylinder can burst, we need to use the ideal gas law equation:

PV = nRT

Where: - P is the pressure of the gas in the cylinder (3000 kPa) - V is the volume of the gas in the cylinder (74.7 L) - n is the number of moles of the gas (which we can calculate using the given mass and molar mass) - R is the ideal gas constant (8.314 J/(mol·K)) - T is the temperature in Kelvin (the temperature we want to find)

First, let's calculate the number of moles of carbon dioxide gas in the cylinder using the given mass and molar mass:

n = mass / molar mass

Given: - Mass of carbon dioxide gas = 1.98 kg - Molar mass of carbon dioxide gas = 0.044 kg/mol

Substituting the values into the equation, we get:

n = 1.98 kg / 0.044 kg/mol

Calculating this, we find that the number of moles of carbon dioxide gas is approximately 45 moles.

Now, let's rearrange the ideal gas law equation to solve for temperature:

T = PV / (nR)

Substituting the given values into the equation, we get:

T = (3000 kPa) * (74.7 L) / (45 mol) * (8.314 J/(mol·K))

Converting the pressure from kPa to Pa and the volume from L to m^3, we get:

T = (3000 * 10^3 Pa) * (74.7 * 10^-3 m^3) / (45 mol) * (8.314 J/(mol·K))

Simplifying the equation, we find:

T ≈ 228.3 K

Therefore, the cylinder can potentially burst at a temperature of approximately 228.3 Kelvin.