It is desired to increase the volume of the 800 ml of a gas by 20% keeping the pressure constant . To what temperature the gas be heated , if the initial temperature is 22 ° C.
Answers
Answer:-
Given:
Initial Volume (V1) = 800 ml
And,
Volume is increased by 20% with increase of temperature.
→ Final Volume (V2) = 800 * (120) / 100 = 960 ml
Initial Temperature (T2) = 22° C = 273 + 22 = 295° K
Let,
Final Temperature (T1) = T K.
We know that,
Charle's law of gases states that at constant pressure the Volume of a gas is directly proportional to it's temperature.
That is,
V ∝ T
→ V/T = k (Constant)
→ T = 354° - 273°
→ T = 81° C
Therefore, the temperature of the gas increases to 81° C when the Volume of the gas is increased by 20%.
Given:
- Initial Volume= V1= 800 ml
- Volume is increased by 20%
- Initial temperature= 22°C = 22+ 273K = 295K
━━━━━━━━━━━━━━━━━━
Increase in Volume
=20%
=
= 160 ml
Therefore Final volume = V2
=800+160ml
=960ml
━━━━━━━━━━━━━━━━━━
Considering the gas as an ideal gas. Apply Charles law.
According to Charles Law at constant pressure the volume of a ideal gas is directly proportional to it's temperature.
The formula is:
Also; 354K
= 354-273 °C
=
━━━━━━━━━━━━━━━━━
Therefore the gas can be heated till T2= 354K= 81°C