## In a given process of an ideal gas, dW = 0 dQ < 0. Then for the gas

Q: In a given process of an ideal gas, dW = 0 dQ < 0. Then for the gas

(a)The temperature will decrease

(b)The volume will increase

(c)The pressure will remain constant

(d)The temperature will increase

Ans: (a)

Sol: From First Law of thermodynamics ;

dQ = dU + dW

if , dW = 0 ; dQ = dU

Since , dQ < 0

$\large U_{final} < U_{initial}$

Hence , temperature will decrease .

## Two monoatomic ideal gases 1 and 2 of moleclar masses m1 and m2 respectively are enclosed in…

Q: Two monoatomic ideal gases 1 and 2 of moleclar masses m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to the gas 2 is given by

(a) $\large \sqrt{\frac{m_1}{m_2}}$

(b) $\large \sqrt{\frac{m_2}{m_1}}$

(c) $\large \frac{m_1}{m_2}$

(d) $\large \frac{m_2}{m_1}$

Ans: (b)

Sol: $\large v = \sqrt{\frac{\gamma R T}{M}}$

$\large v \propto \sqrt{\frac{1}{M}}$

$\large \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{m_2}{m_1}}$

## Starting with the same initial conditions, an ideal gas expands from volume V1 and V2 in three different ways…

Q: Starting with the same initial conditions, an ideal gas expands from volume V1 and V2 in three different ways, the work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic, then

(a)W2> W1> W3

(b)W2> W3> W1

(c)W1> W2> W3

(d)W1> W3> W2

Ans: (a)

Sol: Draw P-V graph for three different processes .

As Area under P-V graph gives work done by the gas.

(Area)2 > (Area)1 > (Area)3

W2 > W1 > W3

## A monoatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston…

Q: A monoatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperatrue T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by

(a) (L1/L2)2/3

(b) (L1/L2)

(c) L2/L1

(d) (L2/L1)2/3

Ans: (d)

$\large T V^{\gamma – 1} = constant$

$\large T_1 V_1^{\gamma – 1} = T_2 V_2^{\gamma – 1}$

For monoatomic gas , γ = 5/3

$\large \frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma – 1}$

$\large \frac{T_1}{T_2} = (\frac{A L_2}{A L_1})^{\frac{5}{3} – 1}$

(Where A = Area of cross-section of piston)

$\large \frac{T_1}{T_2} = (\frac{L_2}{L_1})^{\frac{2}{3}}$

## A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T…

Q: A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

(a) 4 RT

(b) 15 RT

(c) 9 RT

(d) 11 RT

Ans: (d)

Sol: Internal Energy of n moles of an ideal gas at temperature T is

$\large U = n (\frac{f}{2}R T)$

f = degree of freedom

f = 5 for O2 & f = 3 for Ar

$\large U = U_{O_2} + U_{Ar}$

$\large U = 2 (\frac{5}{2}R T) + 4 (\frac{3}{2}R T)$

U = 11 R T