Question

Two moles of an ideal monoatomic gas occupy a volume 2V at temperature 300 K, it expands to a volume 4V adiabatically, then the final temperature of gas is

• A. 179 K
• B. 189 K
• C. 199 K
• D. 219 K

Answer 1

The correct option is B 189 K

Given,

$n=2$

$V_1=2V$

$V_2=4V$

$T_1=300K$

For ideal monoatomic gas, $\gamma =1.66$

In adiabatic process,

$P{V}^{\gamma }=k\left(constant\right)$. . . . . .(1)

From ideal gas equation,

$PV=nRT=2RT$

$P=\frac{2RT}{V}$

From equation (1),

$\frac{2RT}{V}{V}^{\gamma }=k$

$T{V}^{\gamma -1}=k^{\prime }\left(constant\right)$

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

$T_2=T_1(\frac{V_1}{V_2}{)}^{\gamma -1}$

$T_2=300\times (\frac{2V}{4V}{)}^{1.66-1}$

$T_2=300\times 0.630$

$T_2=189K$

The final temperature of gas is $189K$.

The correct option is B.