Question

$\frac{1}{2}$ mole of Helium gas is contained in a container at S.T.P. The heat energy needed to double the pressure of the gas, keeping the volume constant (heat capacity of the gas = $3J{g}^{-1}{K}^{-1}$)is

• A. $3276J$
• B. $1638J$
• C. $819J$
• D. $409.5J$

Answer 1

Answer: (B)

$1638J$

Given :1/212 mole of Helium gas is contained in a container at S.T.P. The heat energy needed to double the pressure of the gas, keeping the volume constant (heat capacity of the gas = 3Jg−1K−13Jg−1K−1)is

Solution :

$C_v=3$ , M=4 then $M{C}_v=12$

$P\propto T$

Then $\frac{P_2}{P_1}=\frac{T_2}{T_1}$ = 2 $\Rightarrow$ $T_2=2T_1$

$\Delta T$= $T_2-T_1$ = $T_1$ =273K

Heat Required $\Delta Q=nM{C}_v\Delta T$

$\Delta Q=\frac{1}{2}\times 12\times 273$ = 1638J

The Correct Opt = B