Question

Газ, при давлении 2*10^5 Па и температуре 15°С равен объёму 5 л. Чему равен объём газа этой массы при нормальных условиях?

Answer 1

p1=2*10^5 Па  р2=10^5 Па   V1=5*10^-3 м3  Т1=273+15=288 К   Т2=273 К   V2=?

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p1*V1/T1=p2*V2/T2        V2=p1*V1*T2/(p2*T1)=2*10^5*5*10^-3*273/(10^5*288)=9.5*10^-3 м3    (9,5 л)

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Answer 2

Gas Volume Calculation

To calculate the volume of the gas at normal conditions, we can use the combined gas law, which states that the ratio of the initial pressure and temperature to the final pressure and temperature is equal to the ratio of the initial volume to the final volume. The formula for the combined gas law is:

P1V1/T1 = P2V2/T2

Where: - P1 = initial pressure - V1 = initial volume - T1 = initial temperature - P2 = final pressure (at normal conditions) - V2 = final volume (what we want to find) - T2 = final temperature (at normal conditions)

Given: - Initial pressure, P1 = 2*10^5 Pa - Initial volume, V1 = 5 L - Initial temperature, T1 = 15°C = 15 + 273 = 288 K - Normal pressure, P2 = 1 atm = 1.013*10^5 Pa - Normal temperature, T2 = 0°C = 273 K

We can rearrange the combined gas law formula to solve for V2:

V2 = (P1 * V1 * T2) / (P2 * T1)

Substitute the given values: V2 = (2*10^5 * 5 * 273) / (1.013*10^5 * 288)

Calculating the value of V2 using the given values: V2 ≈ 11.2 L

So, the volume of the gas at normal conditions is approximately 11.2 liters.