Q: COMPREHENSION:

A quantity of an ideal monoatomic gas consists of n moles initially at temperature T_{1} . The pressure and volume are then slowly doubled in such a manner so as to trace out a straight line on a P-V diagram

1 . For this process, the ratio $\frac{W}{nRT_1}$ is equal to (where W is work done by the gas) :

(A) 1.5

(B) 3

(C) 4.5

(D) 6

2. For the same process, the ratio $\frac{Q}{nRT_1}$ is equal to (where Q is heat supplied to the gas) :

(A) 1.5

(B) 3

(C) 4.5

(D) 6

3. If C is defined as the average molar specific heat for the process then $\frac{C}{R}$ has value

(A) 1.5

(B) 2

(C) 3

(D) 6

Ans: 1:(A) ; 2: (D) ; 3: (B)

Solution: 1: W = Area under the curve $= \frac{3}{2}P_1 V_1$

$P_1 V_1 = n R T_1 $

Therefore $\frac{W}{n R T_1} = \frac{\frac{3}{2}P_1 V_1}{P_1 V_1} $

2: Q = dU + W

dU = n C_{v} dT

For final state P_{2}V_{2} = 2P_{1} 2V_{1} = 4P_{1}V_{1} = nR(4T_{1})

Hence final temp. is 4T_{1}

$dU = n \frac{3}{2}R.3T_1 = \frac{9}{2}n R T_1 $

$Q = \frac{3}{2}n R T_1 + \frac{9}{2}n R T_1 = 6 n R T_1 $

$\frac{Q}{n R T_1} = 6 $

3: n C ΔT = Q

⇒ n C ΔT = 6 n R T_{1}

dT = 4T_{1} – T_{1} = 3T_{1}

n . C . 3 T_{1} = 6 n R T_{1}

$\frac{C}{R} = 2$