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**Question 1: A 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)****(a) K to L and L to M****(b) L to M and N to K****(c) L to M and M to N****(d) 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.**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.litre ^{2} mol^{–2}) is (IIT JEE 2012)**

Van der waals equation for 1 mole of real gas:

(P+a/v

Given that, b=0

So, the equation becomes;

(P+a/v

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 = (y

Hence, the correct option is c.

The unknown compound would not follow the ideal gas equation and thus the problem can

be solved by applying the ideal gas equation to the He gas only.

Using ideal gas equation:

V = (nRT )/P

V = (0.1 × 0.0821 × 273)/(0.32)

V = 7 L

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