At what temperature will 0.3 mitre qube of a gas kept at 290 Kelvin be doubled in volume,if the pressure is kept constant
Answers
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
Answer :
Initial volume of gas = 0.3m³
Initial temperature of gas = 290K
We have to final final temperature of gas at which its volume becomes double
★ Gay Lussac's law : At constant pressure, pressure of a fixed amount of a gas varies directly with the temperature.
Mathematically, V ∝ T
Or we can say that, V/T = V'/T'
- V denotes initial volume
- T denotes initial temperature
- V' denotes final volume
- T' denotes final temperature
By substituting the values, we get
➠ V/T = V'/T'
➠ 0.3/290 = 2(0.3)/T'
➠ T' = 290 × 2
➠ T' = 590 K
∴ Volume of gas becomes double at 590K.
Additional Information :
- Boyle's law : At constant temperature, the pressure of a amount of gas varies inversely with its volume.
- Charles' law : Pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.