Question

$10$ litres of a mono atomic ideal gas at $0^{o}C$ and $10$ $atm$ pressure is suddenly released to $1$ $atm$ pressure and the gas expands adiabatically against this constant pressure. The final temperature and volume of the gas respectively are:

• A. $T=174.9$K, $V=64.0$ liters
• B. $T=153$K, $V=57$ liters
• C. $T=165.4$K, $V=78.8$ liters
• D. $T=161.2$K, $V=68.3$ liters

Answer 1

The correct option is A $T=174.9$K, $V=64.0$ liters

Given that monoatomic gas

$T_1=0^{0}C=273K$, $V_1=10L$, $P_1=10atm$

$P_2=1atm$, $V_2=V$

For an adiabatic process

work, $w=\frac{(P_2{V}_2-P_1{V}_1)}{\gamma -1}$

where $\gamma =5/3$ for monoatomic gas

Also, for expansion against constant P, $w=-P_{ext}(V_2-V_1)$

thus, $-P_{ext}(V_2-V_1)=\frac{(P_2{V}_2-P_1{V}_1)}{\gamma -1}$

$-1(V-10)=\frac{(1.V-10.10)}{5/3-1}$

$10-V=3\times \frac{(V-100)}{2}$

$5V=320$

$V=64L$

Applying ideal gas law:

$P_1.V_1=nR{T}_1$

$10.10=n\times 0.0821\times 273$

$n=4.47$

Also, $P_2.V_2-P_1.V_1-=nR(T_2-T_1)$

$1\times 64-10\times 10=4.47\times 0.0821\times (T_2-273)$

$-36=0.367(T_2-273)$

$T_2=174.9K$

Thus final volume is 64L and temperature is 174.9K.