## Two containers of equal volume contain the same gas at pressure p1 and p2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio p/T is equal to

Q: Two containers of equal volume contain the same gas at pressure p1 and p2 and absolute temperatures T1 and T2 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)p1/T1 + p2/T2
(b)½ [p1/T1 + p2/T2]
(c)[p1T2 + p2T1]/(T1+T2)
(d)[p1T2 – p2T1]/(T1-T2)
Solution : (b)
For a closed system, the total mass of gas or the number of moles remains constant.
P1V = n1RT1 ,     p2V = n2RT2
p(2V) = (n1 + n2)RT
p(2V) = (p1V/RT1 + p2V/RT2)RT
p/T =1/2[p1/T1 + p2/T2]

## 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 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
(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)
U1 = U2
 f1n1T1 = f2n2T2
n1/n2 = f2T2/f1T1 = 3×2/5×1 = 6/5
Here f2 = degrees of freedom of He = 3 and f1 = degrees of freedom of H2 = 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 = nCP (ΔT)A and (ΔQ)B = nCV (ΔT)B
Given that  (ΔQ)A = (ΔQ)B
(ΔT)B = Cp/Cv(ΔT)A
= (1.4) (30)        (Since Cp/Cv = 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
k1(100-θ) = k2(θ-0)
Here , k1/k2 = 2/3
By solving , θ = 40°C

## The radiation force due to source of power P on a perfectly reflecting surface will be

Q: The radiation force due to source of power P on a perfectly reflecting surface will be
(a) P/3c
(b) 2P/c
(c) P/2c
(d) P/c
Solution :(b)
E = m0c2
or , m0c = E/c
Change in momentum Δp = 2m0c = 2E/c
F = dp/dt = 2p/c (As dE/dt =p)