Question

One mole of a non linear triatomic ideal gas is
1 atm.
on linear triatomic ideal gas is expanded adiabatically at 300 K from 16 atm to
Find the final temperature if expansion is carried out reversibly.
(A) 140K
(B) 150K
(C) 160K
(D) 120K​

Answer 1

answer : option (B) 150 K

we know, in adiabatic process

$PV^{\gamma}=\textbf{constant}$

using gas equation, V = nRT/P

$P\left(\frac{nRT}{P}\right)^{\gamma}=\textbf{constant}$

$P^{(1-\gamma)}T^{\gamma}=\textbf{constant}$

$P_1^{(1-\gamma)}T_1^{\gamma}=P_2^{(1-\gamma)}T_2^{\gamma}$

for triatomic molecule, $\gamma$ = 4/3

so, (1atm)^(1 - 4/3)(300K)^(4/3) = (16atm)^(1-4/3)T^(4/3)

⇒300⁴ = (16)^(-1) T⁴

⇒(300/2)⁴ = T⁴

⇒T = 150K

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Answer 2

Explanation:

calculate the vapour pressure of an aqueous solution of ammonia gas glucose 100 degree Celsius