One mole of monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1 L to 2 L. Find out the final temperature (R=0.0821 L atm K -1 mol -1)

Question

One mole of monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1 L to 2 L. Find out the final temperature (R=0.0821 L atm K -1 mol -1) (A) T (B) (T/ (2)(5)3-1) (C) T -(2/ 3 × 0.082) (D) T + (2/ 3 × 0.082)

Tardigrade Admin · Accepted Answer

For an irreversible, adiabatic process; 0=CV(T2-T1)+pe(V2-V1) Substituting the values CV(T-T2)=1(2-1) atm L ⇒ T-T2=(1/CV)=(2/3 R) ⇒ T2=T-(2/3 × 0.082)

Q. One mole of monoatomic ideal gas expands adiabatically at initial temperature $T$ against a constant external pressure of $1 a t m$ from $1 L$ to $2 L$ . Find out the final temperature $(R = 0.0821 L a t m K^{- 1} m o l^{- 1})$

$T$

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

$T - \frac{2}{3 \times 0.082}$

$T + \frac{2}{3 \times 0.082}$

For an irreversible, adiabatic process;
$0 = C_{V} (T_{2} - T_{1}) + p_{e} (V_{2} - V_{1})$
Substituting the values
$C_{V} (T - T_{2}) = 1 (2 - 1)$ atm $L$
$\Rightarrow T - T_{2} = \frac{1}{C _{V}} = \frac{2}{3 R}$
$\Rightarrow T_{2} = T - \frac{2}{3 \times 0.082}$

Q. One mole of monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1atm from 1L to 2L. Find out the final temperature (R=0.0821LatmK−1mol−1)

T

(2)35​−1T​

T−3×0.0822​

T+3×0.0822​

For an irreversible, adiabatic process; 0=CV​(T2​−T1​)+pe​(V2​−V1​) Substituting the values CV​(T−T2​)=1(2−1) atm L ⇒T−T2​=CV​1​=3R2​ ⇒T2​=T−3×0.0822​

Q. One mole of monoatomic ideal gas expands adiabatically at initial temperature $T$ against a constant external pressure of $1 a t m$ from $1 L$ to $2 L$ . Find out the final temperature $(R = 0.0821 L a t m K^{- 1} m o l^{- 1})$

$T$

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

$T - \frac{2}{3 \times 0.082}$

$T + \frac{2}{3 \times 0.082}$

For an irreversible, adiabatic process;
$0 = C_{V} (T_{2} - T_{1}) + p_{e} (V_{2} - V_{1})$
Substituting the values
$C_{V} (T - T_{2}) = 1 (2 - 1)$ atm $L$
$\Rightarrow T - T_{2} = \frac{1}{C _{V}} = \frac{2}{3 R}$
$\Rightarrow T_{2} = T - \frac{2}{3 \times 0.082}$