Two different ideal diatomic gases A and B are initially in the same state. A and B are then expanded to the same final volume through adiabatic and isothermal processes respectively. If $P_{A}, P_{B}$ and $T_{A}, T_{B}$ represents the final pressure and temperature of A and B respectively then

P_{A} < P_{B}

and

T_{A} < T_{B}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P_{A} > P_{B}

and

T_{A} > T_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P_{A} > P_{B}

and

T_{A} < T_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P_{A} < P_{B}

and

T_{A} > T_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $P_{A} < P_{B}$ and $T_{A} < T_{B}$
Initial pressure $= P_{A}$
Initial temperature $= T_{A}$
Final pressure $= P_{B}$
Final temperature $= T_{B}$
$∴$ For same volume, i.e, isochoric process,
$P_{A} < P_{B}$
$T_{A} < T_{B}$

Suggest Corrections

Similar questions

Q. Figure (24 - E3) shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are

p_{A}, T_{A}, V

in the vessel A and

p_{B}, T_{B}, V

in the vessel B. The vessels are now connected through a small tube. The ratio between P and T is

Q. The value of C_p − C_vis 1.00 R for a gas sample in state A and 1.08 R in state B. Let p_A and p_B denote the pressures and T_A and T_B denote the temperatures of the states A and B, respectively. It is most likely that
(a) p_A < p_B and T_A > T_B
(b) p_A > p_B and T_A < T_B
(c) p_A = p_B and T_A < T_B
(d) p_A > p_B and T_A = T_B

Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are $p_{A}, T_{A}$ , V in the vessel A and $p_{B}, T_{B}$ , V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy $\frac{p}{T} = \frac{1}{2} (\frac{p_{A}}{T_{A}} + \frac{p_{B}}{T_{B}})$ when equilibrium is achieved.