Question

An ideal monoatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure?

Answer 1

Data: T_f = 2T_i, monoatomic gas ∴ ϒ = 5/3

P_iV_iϒ = P_fV_fϒ in an adiabatic process

Now, PV = ηRT ∴ V = `(η"RT")/"P"`

∴ V_i = `(η"RT"_"i")/"P"_"i"` and V_f = `(η"RT"_"f")/"P"_"f"`

∴ P_i `((η"RT")/"P"_"i")^ϒ="P"_"f"((η"RT"_"f")/"P"_"f")^ϒ`

∴ `"P"_"i"^(1-ϒ)"T"_"i"^ϒ="P"_"f"^(1-ϒ)"T"_"f"^ϒ`

∴ `(("T"_"f")/"T"_"i")^ϒ=(("P"_"i")/("P"_"f"))^(1-ϒ)`

∴ `(("T"_"f")/"T"_"i")^ϒ=(("P"_"f")/("P"_"i"))^(ϒ-1)`

∴ 2^5/3 = `(("P"_"f")/("P"_"i"))^(5//3-1)=(("P"_"f")/("P"_"i"))^(2//3)`

∴ `5/3` log 2 = `2/3` log `"P"_"f"/"p"_"i"`

∴ `5/3xx0.3010 = 2/3` log `(("P"_"f")/"P"_"i")`

∴ (2.5) (0.3010) = log `(("P"_"f")/("P"_"i"))`

∴ 0.7525 = log `("P"_"f"/"P"_"i")`

∴ `("P"_"f")/"P"_"i"` = antilog 0.7525 = 5.656

This is the ratio of the final pressure (Pf) to the initial pressure (Pi).

Answer 2

Given: T_f = 2T_i, `"P"_"f"/"P"_"i"` = ?

Formula: 

`"P"_"f"/"P"_"i" = (("T"_"f")/("T"_"i"))^(gamma/(gamma - 1))`

∴ `"P"_"f"/"P"_"i" = ((2"T"_"i")/("T"_"i"))^((5"/"3)/(5"/"3 - 1))` ............`(∵ "For mono-atomic gas", gamma = 5/3)`

∴ `"P"_"f"/"P"_"i" = 2^2.5`

∴ `"P"_"f"/"P"_"i" = 5.6568`