Question

Какой объем занимают 400 г кислорода при давлении 2*10^6 Па и температуре 27 С

Answer 1

m=400г

P=2*10^6 Па

T=27 C = 300 K

R=8.31

 

PV = m/M*RT

V=m/MP*RT =400/(32*2*10^6) *8.31*300 =0.0156 м3 =15.6 л

Answer 2

Calculating the Volume of 400g of Oxygen

To calculate the volume of 400g of oxygen at a pressure of 2*10^6 Pa and a temperature of 27°C, we can use the ideal gas law formula:

V = (nRT)/P

Where: - V = volume - n = number of moles - R = ideal gas constant - T = temperature in Kelvin - P = pressure

First, we need to calculate the number of moles of oxygen using the formula:

n = (mass of substance)/(molar mass of substance)

The molar mass of oxygen (O2) is approximately 32 g/mol.

Then, we convert the temperature from Celsius to Kelvin using the formula:

T(K) = T(°C) + 273.15

Let's calculate the volume using the given values.

Calculation

1. Calculate the number of moles of oxygen: - n = (400g)/(32g/mol) - n ≈ 12.5 mol 2. Convert the temperature to Kelvin: - T(K) = 27°C + 273.15 - T(K) ≈ 300.15 K 3. Use the ideal gas law formula to calculate the volume: - V = (nRT)/P - V = (12.5 mol * 8.314 J/(mol*K) * 300.15 K) / (2*10^6 Pa) - V ≈ 0.1875 m^3

Conclusion

The volume of 400g of oxygen at a pressure of 2*10^6 Pa and a temperature of 27°C is approximately 0.1875 cubic meters.