One mole of an ideal gas undergoes a cyclic p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a cyclic process on a $P-V$ diagram,the work done is equal to the area enclosed by the cycle.The cycle is a rectangle with vertices $1, 2, 3,

Vedclass · Accepted Answer

$R(\sqrt{T_3}-\sqrt{T_1})^2$

Vedclass · Answer

(B) For a cyclic process on a $P-V$ diagram,the work done is equal to the area enclosed by the cycle.The cycle is a rectangle with vertices $1, 2, 3, 4$.Let the coordinates be $(P_1, V_1), (P_2, V_1), (P_2, V_2), (P_1, V_2)$.The work done $W = (P_2 - P_1)(V_2 - V_1)$.From the ideal gas law $PV = RT$ (for $n=1$ mole):At point $1$: $P_1V_1 = RT_1$At point $3$: $P_2V_2 = RT_3$Since points $2$ and $4$ lie on the same isotherm,$T_2 = T_4 = T_0$.At point $2$: $P_2V_1 = RT_0$At point $4$: $P_1V_2 = RT_0$Thus,$P_2V_1 = P_1V_2 \implies P_2/P_1 = V_2/V_1 = k$.Then $P_2 = kP_1$ and $V_2 = kV_1$.$P_1V_1 = RT_1$$P_2V_2 = k^2 P_1V_1 = k^2 RT_1 = RT_3 \implies k^2 = T_3/T_1 \implies k = \sqrt{T_3/T_1}$.$W = (P_2 - P_1)(V_2 - V_1) = P_1(k-1) V_1(k-1) = P_1V_1(k-1)^2$.$W = RT_1(\sqrt{T_3/T_1} - 1)^2 = RT_1(\frac{\sqrt{T_3} - \sqrt{T_1}}{\sqrt{T_1}})^2 = R(\sqrt{T_3} - \sqrt{T_1})^2$.

One mole of an ideal gas undergoes a cyclic process,consisting of two isochores and two isobars. Temperatures at points $1$ and $3$ are $T_1$ and $T_3$ respectively. Find the work done by the gas over the cycle,if points $2$ and $4$ lie on the same isotherm.

Similar Questions

There are two samples $A$ and $B$ of a certain gas,which are initially at the same temperature and pressure. Both are compressed from volume $V$ to $\frac{V}{2}$. Sample $A$ is compressed isothermally while sample $B$ is compressed adiabatically. The final pressure of $A$ is

Check whether the following statements are true or false:
$1.$ The change in internal energy $\Delta U = 0$ in a cyclic process.
$2.$ In an adiabatic process,temperature remains constant.
$3.$ The internal energy of a system during an isothermal process decreases.

The slopes of isothermal and adiabatic curves are related as

Heat is applied to a rigid diatomic gas at constant pressure. The ratio $\Delta Q : \Delta U : \Delta W$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module