Question

PV=mRT

Where R=0.287 KJ/Kg K

Mass=2 kg, T=35 degrees C, in cubic container with inside demensions of 0.750 meters each side, find pressure in Pa

Answer 1

The pressure of gas in a 0.75-meter cubic container can be calculated using the ideal gas law. When given mass, temperature, and the specific gas constant, we find that the pressure is 993470 Pa.

Step-by-step explanation:

The question involves using the ideal gas law, PV=mRT, to find the pressure of a gas. The volume of the cubic container can be calculated as V = 0.750m * 0.750m * 0.750m. To find the pressure, the temperature T must first be converted to Kelvin: T(K) = 35°C + 273.15 = 308.15K. Now we can solve for pressure P using the provided mass, temperature, specific gas constant R, and calculated volume:

P = (m*R*T) / V

Substituting the known values we get:

P = (2kg * 0.287 kJ/kg·K * 308.15 K) / (0.421875 m³)
= (0.574 kJ/kg·K * 308.15 K) / (0.421875 m³)
= 419.13825 kJ/m³ / 0.421875 m³
= 993.47 kPa

To convert kPa into Pa, we multiply by 1000:

Pressure (P) = 993.47 kPa * 1000 Pa/kPa
= 993470 Pa

Answer 2

The answer to your question is: P = 419933.5 Pa

Step-by-step explanation:

Data

R = 0.287 KJ/Kg°K = 287 J/Kg°K

mass = 2 kg

T = 35 °C = 308°K

Container = 0.750 meters each side

Volume = l x l x l

= 0.750 x 0.750 x 0.750

= 0.421 m³

P = mRT / V

P = (2)(287)(308) / 0.421

P = 176792 / 0.421

P = 419933.5 Pa