Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

One mole of monoatomic ideal gas at T (K) is expanded from 1 L to 2 L adiabitcally under constant external pressure of 1 atm. The final temperature of gas in kelvin is

(a) T
(b) $\frac{T}{2^{5/3-1}}$
(c) T - $\frac{2}{3\times 0.0821}$
(d) T + $\frac{2}{3\times 0.0821}$

Can you answer this question?

1 Answer

0 votes
Answer: T - $\frac{2}{3\times 0.0821}$
Here $P_{ext} = 1 atm$
$V_1$ = 1 L and $V_2$ = 2 L
R = 0.0821 L atm $K^{-1}mol^{-1}$
$\gamma = 5/3$ (For a monoatomic gas)
$T_! = T$ K and $T_2=?$
Now, $C_P - C_V=R$ or $\frac{C_P}{C_V}- 1=\frac{R}{C_V}$
$\Rightarrow \gamma - 1=\frac{R}{C_V}$
$C_V = \frac{R}{\gamma -1}$
$\Rightarrow C_V(T_2-T_1) = P_{ext}(V_1-V_2)$
$\Rightarrow \frac{R}{\gamma -1}(T_2-T_1) = P_{ext} (V_1 - V_2)$
Substituting the values,
$\frac{0.0821}{5/3-1}(T_2-T_1) = 1\times (1-2)$
$\Rightarrow T_2-T=\frac{2/3}{0.0821}$
$\Rightarrow T_2 = T - \frac{2}{3\times 0.0821}$
answered Feb 28, 2014 by mosymeow_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App