Physics Thermodynamics

During the process, area under P – V graph continuously increases but rate of increment varies.

If $P_1=P$ then $P_2=P+5\%$ of $P=1.05P.$

From Boyle’s law, $PV=constant$ therefore,

$\frac{V_2}{V_1}=\frac{P_1}{P_2}=\frac{P}{1.05P}=\frac{100}{105}$

Fractional change in volume is

$\frac{\Delta V}{V}=\frac{V_2-V_1}{V_1}=\frac{100-105}{105}=-\frac{5}{105}$

Therefore, percentage change in volume is

$\frac{\Delta V}{V}\times 100\%=-\frac{5}{105}\times 100\%=-4.76\%$

That is, volume decreases by $4.76\mathrm{\%}.$

From first law of thermodynamics

$?W=?Q-?U$

$?Q=n{C}_p?T$

$?U=n{C}_v?T$

$?W=n(C_p-C_v)?T$

$\frac{?Q}{?W}=\frac{\gamma }{\gamma -1}\left. [\frac{C_p}{C_p}=\gamma \right]$

$COP=\frac{Q_t}{W}=\frac{T_{low}}{T_{high}-T_{low}}=\frac{250}{300-250}=5$

$\frac{Q_L}{W}=5$

$W=\frac{500*4.2}{5}420J$

From first law of thermodynamics

$?W=?Q-?U$

$?Q=n{C}_p?T$

$?U=n{C}_v?T$

$?W=n(C_p-C_v)?T$

$\frac{?Q}{?W}=\frac{\gamma }{\gamma -1}\left. [\frac{C_p}{C_p}=\gamma \right]$

$\eta =1-\frac{T_2}{T_1}=1-\frac{500}{1000}=\frac{1}{2}$

$\eta =\frac{W}{Q_1}$

$\Rightarrow Q_1=\frac{W}{\eta }=\frac{1000}{1/2}$

$\Rightarrow Q_1=2000J$

During the process are under P – V graph continuously increases but rate of increment varies.

$R\propto \left(\theta -{\theta }_0\right)$

$\Rightarrow lo{g}_{e}R=lo{g}_{e}k+lo{g}_e\left(\theta -{\theta }_0\right)$