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  • An ideal gas is taken from the state A(p, V) to the state along a st

An ideal gas is taken from the state A(p, V) to the state\mathrm{B(p / 2,2 V)} along a straight line path as shown in figure. select the correct statement from the following

Option: 1

Work done by the gas in going from A to B exceeds the work done in going from A to B under isothermal conditions


Option: 2

In the T − V diagram, path AB would become a hyperbola,


Option: 3

In the p − T diagram, path AB would be part of hyperbola


Option: 4

In going from A to B, the temperature T of gas first increases to a maximum value 1 and then decreases


Answers (1)

best_answer

Isothermal curve from A to B will be parabolic with lesser area under the curve than
the area under straight line AB. Therefore, work done by the gas in going straight from
A to B is more. Temperature (a) in correct
If \mathrm{p_0, V_0} be the intercepts of curve on p and V axes, then its equation is obtained from

\mathrm{\begin{aligned} & y=m x+c \\ & \text { ie } p=\frac{p_0}{V_0} V+p_0 \\ & \text { or } \frac{R T}{V}=\frac{p_0 V}{V_0}+p_0 \\ & \text { or } T=\frac{p_0}{V_0 R} V^2+\frac{p_0 V}{R} \end{aligned}}

Which is the equation of a parabola. Hence T − V curve is parabolic. Therefore (b) is
correct.
Also\mathrm{(p / 2) \times(2 V)=p V=} constant ie process is isothermal

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avinash.dongre

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