91影视

• No. of mole of gas are$n=2\text{鈥尘辞濒}$.
• Pressure is$p=1\text{鈥塧迟尘}$.
• Temperature is $T=300\text{鈥块}$.

The expression for the$纬$ is given by,

$纬=\frac{C_P}{C_V}$ ...... (i)

Here$C_P$ is the specific heat capacity at constant pressure,$C_V$ is the specific heat capacity at constant volume.

For diatomic gas,

$C_P=\frac{7}{2}R$and

$C_V=\frac{5}{2}R$

Substitute $\frac{7}{2}R$for$C_P$and $\frac{5}{2}R$for $C_V$into the equation (i),

$\begin{array}{c}纬=\frac{C_p}{C_v}\\ =\frac{\left(\frac{7}{2}\right)R}{\left(\left(\frac{5}{2}\right)R\right)}\\ =\frac{7}{5}\end{array}$

$鈬�纬=1.4$

We can find the volume of the gas

$1\text{鈥塧迟尘}=1.01脳{10}^5\text{鈥塒补}$

We can find the volume of gas using the gas law

$$\begin{gathered} pV=nRT \\ 鈬�V=\frac{nRT}{p} \end{gathered}$$

Substitute $8.3145\text{鈥�}J/mol\text{鈥�}鈰�K$ for R, $2\text{鈥�}mol$for n,$300\text{鈥�}K$ for T , $1.01脳{10}^5\text{鈥塒补}$for p into the above equation,

$\begin{array}{c}V=\frac{(2脳8.31脳300)}{1.01脳{10}^5}\\ =0.049{\text{鈥尘}}^3\end{array}$

For an adiabatic process,

$p{V}^{纬}=a$

Substitute$1.01脳{10}^5\text{鈥塒补}$ for p, $0.049{\text{鈥尘}}^3$for V and $1.4$for$纬$ into the above equation,

$\begin{array}{c}\text{a}=1.01脳{10}^5脳{0.049}^{1.4}\\ 鈮�1.5脳{10}^3{\text{N.m}}^2\end{array}$

Therefore the value of a is $1.5脳{10}^3{\text{Nm}}^2$.