1
JEE Main 2016 (Online) 10th April Morning Slot
+4
-1 Initially, the root mean square (rms) velocity of N2 molecules at certain temperature is $$u$$. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new rms velocity will be :
A
u/2
B
2u
C
4u
D
14u
2
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1 At very high pressures, the compressibility factor of one mole of a gas is given by :
A
$${{pb} \over {RT}}$$
B
1 + $${{pb} \over {RT}}$$
C
1 $$-$$ $${{pb} \over {RT}}$$
D
1 $$-$$ $${b \over {\left( {VRT} \right)}}$$
3
JEE Main 2016 (Offline)
+4
-1 Two closed bulbs of equal volume $$(V)$$ containing an ideal gas initially at pressure $${p_i}$$ and temperature $${T_1}$$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $${T_2}.$$ The final pressure $${p_f}$$ is : A
$$2{p_i}\left( {{{{T_2}} \over {{T_1} + {T_2}}}} \right)$$
B
$$2{p_i}\left( {{{{T_1}{T_2}} \over {{T_1} + {T_2}}}} \right)$$
C
$${p_i}\left( {{{{T_1}{T_2}} \over {{T_1} + {T_2}}}} \right)$$
D
$$2{p_i}\left( {{{{T_1}} \over {{T_1} + {T_2}}}} \right)$$
4
JEE Main 2014 (Offline)
+4
-1
If Z is a compressibility factor, van der Waals equation at low pressure can be written as:
A
Z = 1 + $$RT \over Pb$$
B
Z = 1 - $$a \over VRT$$
C
Z = 1 - $$Pb \over RT$$
D
Z = 1 + $$Pb \over RT$$
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