The temperature of an ideal gas increases from 20C to 40C while the pressure stays the same what happens to the volume of the gas
Answers
Answer:
The volume at 40C is 0.936 times that at 20C
Explanation:
We know pv=nRt for ideal gases
Now t₁ = 20C = 293 K t₂ = 40C = 313 K
p₁ = p₂= p R= 8.3145 J/mol
let no of moles stay constant
Now,
p₁v₁=nRt₁
pv₁=nR*(293K)
v₁= [nR*(293K)] / p - (1)
And
p₂v₂=nRt₁
pv₂=nR*(313K)
v₂= [nR*(313K)] / p - (2)
∴ Ratio of v₁ to v₂ ⇒ (1) / (2)
⇒ v₁/v₂ ={ [nR*(293K)] / p} / { [nR*(313K)] / p }
⇒ v₁/v₂ = 293/313
⇒ v₁/v₂ = 0.936
⇒ v₁ = 0.936 v₂
NOTE- I didn't use gas constant as we don't need to find exact value of pressure
Hope this helps
Answer:
The final volume of an ideal gas increases by 1.068 times the initial volume.
Explanation:
Here, the pressure of an ideal gas is constant and also the number of moles of an ideal gas is constant. So, Charle's law is applied.
Charle's law
Charle's law states that the volume of an ideal gas is directly proportional to the temperature of an ideal gas if the pressure and number of moles of an ideal gas remains constant.
Or Initial volume/Final volume = Initial temperature/Final temperature
Let the initial temperature of an ideal gas is given by T, final temperature of an ideal gas is given by T', initial volume of an ideal gas is given by V and final volume of an ideal gas is given by V'.
Temperatures are given as initial temperature = T = 20 °C
Final temperature = T' = 40 °C
Converting the °C temperature to SI units that is in Kelvin by adding 273 in them.
T = 20 + 273 = 293 K
T' = 40 + 273 = 313 K
Now, by applying Charle's law,
V/V' = T/T'
V/V' = 293/313
V'/V = 313/293
V'/V = 1.068
V' = 1.068 V
So, the final volume becomes 1.068 times the initial volume.