## For a gas, the difference between the two principal specific heats is 4150 J/kg K. What is the specific heat…

Q: For a gas, the difference between the two principal specific heats is 4150 J/kg K. What is the specific heat of the gas at constant volume if the ratio of specific heats is 1.4 ?

(a) 5186 J/kg K

(b) 10375 J/kg K

(c) 1660 J/kg K

(d) 8475 J/kg K

Ans:(b)

Sol: Cp-Cv = 4150

Cp/Cv = 1.4,

Cp = 1.4 Cv

1.4 Cv – Cv = 4150

0.4 Cv = 4150

Cv = 4150/0.4

= 10375 J/kg K

## Two metal rods of the same length and are of cross – section are fixed end to end between rigid supports…

Q: Two metal rods of the same length and are of cross – section are fixed end to end between rigid supports. The materials of the rods have Young’s moduli Y1 and Y2 , and coefficients of linear expansion  α1 and α2. The junction between the rods does not shift, if the rods are cooled, when

(a) Y1 α1 = Y2 α2

(b) Y12 α1 =Y22 α2

(c) Y1 α2 = Y2 α1

(d) Y1 α12 = Y2 α22

Ans:(a)

Sol: As the junction between the rods does not shift, tension in both the rods must be the same.

$\displaystyle Y = \frac{Stress}{Strain}$

$\displaystyle = \frac{T/A}{\Delta l/l}$

$\displaystyle Y = \frac{T/A}{(l \alpha \Delta \theta)/l}$

T = Y A α Δθ

As , T1 = T2

Y1 A α1 Δθ = Y2 A α2 Δθ

Y1 α1 = Y2 α2

## 1 cm^3 of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671 cm^3…

Q: 1 cm3 of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671 cm3. If the atmospheric pressure = 1.013 ×105 N/m2 and the mechanical equivalent of heat = 4.19 J/cal., the energy spent in this process in overcoming intermolecular forces is

(a) 540 cals.

(b) 40 cals.

(c) 500 cals.

(d) zero

Ans: (c)

Sol: According to First Law of Thermodynamics

dQ = dU + dW

dU = dQ-dW

= mL – P(dV)

$\displaystyle 1\times 540 – \frac{1.013 \times 10^5 (1671 – 1)\times 10^{-6}}{4.2}$

= 540-40

= 500 cals

## A gas at pressure p is adiabatically compressed so that its density becomes twice that of initial value…

Q: A gas at pressure p is adiabatically compressed so that its density becomes twice that of initial value. Given that γ = Cp/Cv = 7/5 , What will be the final pressure of the gas ?

(a) 2 p

(b) 7/5 p

(c) 2.63 p

(d) p

Ans:(c)

Sol: As , $\displaystyle P_2 V_2^\gamma = P_1 V_1^\gamma$

$\displaystyle P_2 = P_1 (\frac{V_1}{V_2})^\gamma$

$\displaystyle P_2 = P_1 (\frac{\rho_2}{\rho_1})^\gamma$

$\displaystyle P_2 = P_1 (\frac{2 \rho_1}{\rho_1})^\gamma$

$\displaystyle P_2 = p (\frac{2}{1})^{7/5}$

P2 = 2.63 p

## A cylinder fitted with a piston contains 0.2 mole of air at temperature 27 °C. The piston is pushed so slowly…

Q: A cylinder fitted with a piston contains 0.2 mole of air at temperature 27 °C. The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surrounding s. Find the approximate work done by the system if the final volume is twice initial volume.

(a) 600 J

(b) 543 J

(c) 453 J

(d) 344 J

Ans:(d)

Sol: Here, n = 0.2 ,

T1 = 27 °C = (27+273)K = 300 K

V2 = 2 V1 , W = ?
The changes are isothermal

W = 2.303 nRT log10⁡(V2/V1 )

= 2.303 × 0.2 ×8.3 × 300 log10⁡2

= 0.46 × 8.3 × 300 × 0.3010

W = 344 J