(a) T

(b) $\frac{T}{2^{5/3-1}}$

(c) T - $\frac{2}{3\times 0.0821}$

(d) T + $\frac{2}{3\times 0.0821}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

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,

$\frac{0.0821}{5/3-1}(T_2-T_1) = 1\times (1-2)$

$\Rightarrow T_2-T=\frac{2/3}{0.0821}$

$\Rightarrow T_2 = T - \frac{2}{3\times 0.0821}$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...