Question

V=20•10 -3м3 m= 2кг t=27*C T=300k М=2•10 -3кг/моль Р=?

Answer 1

Ну тут задача чисто на уравнение Менделеева-Клайперона pV=v(ню)RT, где v(ню)=m/M. Нам всё известно(R=8,31), сразу подставляем p=(2*8,31*300)/2*20*10^-6=4986/40*10^-6=124650000 Па, или 124,65 МПа

Answer 2

To calculate the pressure (P) of the gas, we can use the Ideal Gas Law, which states:

PV = nRT

where P is the pressure of the gas in Pascals (Pa), V is the volume of the gas in cubic meters (m³), n is the number of moles of gas, R is the ideal gas constant (8.31 J/mol·K), and T is the temperature of the gas in Kelvin (K).

First, let's calculate the number of moles of gas (n) using the mass of the gas (m) and its molar mass (M):

n = m/M

n = 2 kg / (2x10^-3 kg/mol) = 1000 mol

Next, let's convert the temperature from Celsius to Kelvin:

T = 27°C + 273.15 = 300.15 K

Now we can plug in the values we have into the Ideal Gas Law:

PV = nRT

P = nRT/V

P = (1000 mol) x (8.31 J/mol·K) x (300.15 K) / (20x10^-3 m³)

P = 124575 Pa or 124.575 kPa

Therefore, the pressure of the gas is approximately 124.575 kPa.