## A quantity of an ideal monoatomic gas consists of n moles initially at temperature T1 …..

Q: COMPREHENSION:
A quantity of an ideal monoatomic gas consists of n moles initially at temperature T1 . 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

Click to See Solution :
Ans: 1:(A) ; 2: (D) ; 3: (B)

Sol: 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 Cv dT
For final state P2V2 = 2P1 2V1 = 4P1V1 = nR(4T1)
Hence final temp. is 4T1

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

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

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

3: n C ΔT = Q

⇒ n C ΔT = 6 n R T1

dT = 4T1 – T1 = 3T1

n . C . 3 T1 = 6 n R T1

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

## A gas undergoes an adiabatic process and an isothermal process. The two processes are plotted on a P-V diagram….

Q: A gas undergoes an adiabatic process and an isothermal process. The two processes are plotted on a P-V diagram. The resulting curves intersect at a point P. Tangents are drawn to the two curves at P. These make angles of 135º & 121º with the positive V-axis. If tan 59º = 5/3, the gas is likely

(A) monoatomic

(B) diatomic

(C) triatomic

(D) a mixture of monoatomic & diatomic gases

Click to See Solution :
Ans: (A)

Sol: The slope of isothermal curve at point of intersection is

$\displaystyle \frac{dP}{dV} = -\frac{P}{V} = tan135^o$ …(i)

The slope of adiabatic curve at point of intersection is

$\displaystyle \frac{dP}{dV} = -\gamma \frac{P}{V} = tan121^o$ …(ii)

From (i) & (ii)

from (1) and (2)
γ = tan 59° = 1.66 = 5/3
Hence , gas is monoatomic

## An ideal gas undergoes a cyclic process abcda which is shown by pressure density curve….

Q: An ideal gas undergoes a cyclic process abcda which is shown by pressure density curve.

(A) Work done by the gas in the process ‘bc’ is zero

(B) Work done by the gas in the process ‘cd’ is negative

(C) Internal energy of the gas at point ‘a’ is greater than at state ‘c’

(D) Net work done by the gas in the cycle is negative.

Click to See Solution :
Ans: (A),(B)
Sol: $\frac{P}{\rho} = \frac{R}{M_0}T$

Slope of the curve ∝ Temperature

Hence cd and ab are isothermal processes.

i.e. bc and da are constant volume process
(A) and (B) are true. Temp. in cd process is greater than ab. Net work done by the gas in the cycle is
negative, as is clear by the PV-diagram.

$\frac{P}{\rho} = \frac{R}{M_0}T$

## In the figure shown the pressure of the gas in state B is

Q: In the figure shown the pressure of the gas in state B is:

(a) $\frac{63}{25}P_0$

(b) $\frac{73}{25}P_0$

(c) $\frac{48}{25}P_0$

(d) None of these

Click to See Solution :
Ans: (b)

Sol: $AN = 3V_0 cos37^o$

$P_B = \frac{P_0}{V_0}(V_0 + 3 V_0 \times \frac{16}{25})$

$= P_0 (1 + \frac{48}{25}) = (\frac{73}{25})P_0$