Question

$1$ mole of an ideal gas $(C_V=12.55J{K}^{-1}{mol}^{-1})$ at $300K$ compressed adiabatically and reversibly to one-fourth of its original volume. What is the final temperature of the gas?

• A. $750K$
• B. $752K$
• C. $754K$
• D. $756K$

Answer 1

Answer: (B)

$752K$

Number of moles $n=1.$

$C_v=12.55J{K}^{-1}mo{l}^{-}1$

Initial temperature$=T_1=300K$

$C_P=Cv+R=(12.55+8.314)J{K}^{-1}mo{l}^{-1}=20.864J{K}^{-1}mo{l}^{-1}$

adiabatically and reversible compression to $\frac{1}{4}$ of its original volume.

$⟹V_1=V$ $V_2=\frac{V}{4}$

$T{V}^{\gamma -1}=constant\left(T_1\right){\left(V_1\right)}^{\gamma -1}=\left(T_2\right){\left(V_2\right)}^{\gamma -1}\left(300\right){\left(V\right)}^{\gamma -1}=\left(T_2\right){\left(\frac{V}{4}\right)}^{\gamma -1}{T}_2=300\times 4^{0.66}=748.99K\approx 750K$ Value of $\gamma ,\gamma =\frac{C_p}{C_v}\gamma =\frac{20.864}{12.55}\gamma =1.66$

The final temperature of gas is $750K$.