This chapter is currently out of syllabus
1
JEE Main 2016 (Offline)
+4
-1
Out of Syllabus
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
$$2{p_i}\left( {{{{T_1}} \over {{T_1} + {T_2}}}} \right)$$
2
JEE Main 2014 (Offline)
+4
-1
Out of Syllabus
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 + $$Pb \over RT$$
3
JEE Main 2013 (Offline)
+4
-1
Out of Syllabus
For gaseous state, if most probable speed is denoted by C*, average speed by $$\mathop C\limits^{\_\_}$$ and mean square speed by C, then for a large number of molecules the ratios of these speeds are:
A
C*: $$\mathop C\limits^{\_\_}$$ : C = 1.128 : 1.225 : 1
B
C*: $$\mathop C\limits^{\_\_}$$ : C = 1.225 : 1.128 : 1
C
C*: $$\mathop C\limits^{\_\_}$$ : C = 1 : 1.225 : 1.128
D
C*: $$\mathop C\limits^{\_\_}$$ : C = 1 : 1.128 : 1.225
4
AIEEE 2012
+4
-1
Out of Syllabus
The compressibility factor for a real gas at high pressure is :
A
1 + RT/pb
B
1
C
1 + pb/RT
D
1–pb/RT
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