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Question: 1A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure

The pair of isochoric processes among the transformation of states is (IIT JEE-2013)

1) K to L and L to M

2) L to M and N to K

3) L to M and M to N

4) M to N and N to K

Answer: b

Solution: Plot for the processes N to K and L to M are a straight line. This indicates that the value of V remains same at all the points during these processes. These processes take place at constant volume, so these are isochoric processes.

Hence, the correct option is b.

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Question: 2 For one mole of a van der Waals gas when b = 0 and T= 300 K, the PVvs. 1/V plot is shown below. The value of the van der Waals constant ‘a’ (atm.litre2mol–2) is  (IIT JEE 2012)

 

1) 1.0

2) 4.5

3) 1.5

4) 3.0

Answer: C

Solution:

Van der waals equation for 1 mole of real gas:

(P+a/v2)(v-b) =RT

Given that, b=0

So, the equation becomes;

(P+a/v2)v =RT

or

pv + a/v = RT

or

pv = -a×1/v +RT

Comparing with equation of straight line: y = mx +c, it can be concluded that

 pv vs 1/v graph would be a straight line with a negative slope equal to a.

Thus, slope of the plot = a = (y2-y1)/(x2-x1) = (21.6-20.1)/(3-2) =1.5

Hence, the correct option is C.

Question:3 To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol of an unknown...

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