Q: Two containers of equal volume contain the same gas at pressure p1 and p2 and absolute temperatures T_{1} and T_{2} respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio p/T is equal to

(a)p_{1}/T_{1} + p_{2}/T_{2}

(b)½ [p_{1}/T_{1} + p_{2}/T_{2}]

(c)[p_{1}T_{2} + p_{2}T_{1}]/(T_{1}+T_{2})

(d)[p_{1}T_{2} – p_{2}T_{1}]/(T_{1}-T_{2})

Solution : (b)

For a closed system, the total mass of gas or the number of moles remains constant.

P_{1}V = n_{1}RT_{1} , p_{2}V = n_{2}RT_{2}

p(2V) = (n_{1} + n_{2})RT

p(2V) = (p_{1}V/RT_{1} + p_{2}V/RT_{2})RT

p/T =1/2[p_{1}/T_{1} + p_{2}/T_{2}]

# Author: Rajesh Jha

## Internal energy of n1 moles of hydrogen at temperature T is equal to the internal energy of n2 moles of helium at temperature 2T. Then the ratio n1/n2 is

Q: Internal energy of n_{1} moles of hydrogen at temperature T is equal to the internal energy of n_{2} moles of helium at temperature 2T. Then the ratio n1/n2 is

(a)3/5

(b)2/3

(c)6/5

(d)3/7

Solution :(c)

Internal energy of n moles of an ideal gas at temperature T is given by:

U = ½ fnRT (f = degrees of freedom)

U_{1} = U_{2}

f_{1}n_{1}T_{1} = f_{2}n_{2}T_{2}

n_{1}/n_{2} = f_{2}T_{2}/f_{1}T_{1} = 3×2/5×1 = 6/5

Here f_{2} = degrees of freedom of He = 3 and f_{1} = degrees of freedom of H_{2} = 5

## Two cylinders fitted with pistons contains equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of gas in B is

Q: Two cylinders fitted with pistons contains equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of gas in B is

(a) 30 K

(b) 18 K

(c) 50 K

(d) 42 K

Solution :(d)

In cylinder A heat is supplied at constant pressure while in cylinder B heat is supplied at constan volume.

(ΔQ)_{A} = nC_{P} (ΔT)_{A} and (ΔQ)_{B} = nC_{V} (ΔT)_{B}

Given that (ΔQ)_{A} = (ΔQ)_{B}

(ΔT)_{B} = C_{p}/C_{v}(ΔT)_{A}

= (1.4) (30) (Since C_{p}/C_{v} = 1.4)

= 42 K

## Two cylinders P and Q have the same length and diameter and are made of different materials having thermal conductivities in the ratio 2 : 3. These two cylinders are combined to make a cylinder. One end of P is kept at 100° C and another end of Q at 0°C. The temperature at the interface of P and Q is

Q: Two cylinders P and Q have the same length and diameter and are made of different materials having thermal conductivities in the ratio 2 : 3. These two cylinders are combined to make a cylinder. One end of P is kept at 100° C and another end of Q at 0°C. The temperature at the interface of P and Q is

(a) 30°C

(b) 40°C

(c) 50°C

(d) 60°C

Solution : (b) Let temperature at interface is θ

k_{1}(100-θ) = k_{2}(θ-0)

Here , k_{1}/k_{2} = 2/3

By solving , θ = 40°C