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# phy mock questions

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### 1 Answer

- 天同Lv 78 years agoFavorite Answer
1. Cold air has a high density and hence sinks to the floor. Hot air rises and is cooled down by the air-conditioner. Thus a convection current is set up.

2. (a) Use ideal gas equation: PV = nRT

n = PV/RT

No. of moles of gas in partition A, n1 = P(20A)/R(300)

No of moles of gas in partition B, n2 = P(30A)/R(300)

where A is the cross-sectional area of the container

Hence, n1/n2 = 20/30 = 2/3

Since mass of gas is proportional to the no. of moles,

ratio of gas masses = 2:3

(b) Use: P = (1/3).d.c^2

where d is the gas density, and c is the rms speed

Before the gas is heated, Po = (1/3)d(co)^2

After the gas temperature is raised to 477'C (= 750 K)

P = (1/3).d(c)^2

hence, P/Po = (c/co)^2

By Pressure Law, Po/300 = P/750

i.e. P/Po = 750/300

Thus, (c/co)^2 = 750/300

c/co = square-root(750/300) = 1.58

(c) New pressure at partition A, Pa = (n1)R(300)/(L1)A

New pressure at partition B, Pb = (n2)R(750)/(L2)A

where L1 and L2 are the new lengths of partitions A and B respectively

When the piston is in the new equilibrium position, Pa = Pb

(n1)R(300)/(L1)A = (n2)R(750)/(L2)A

(L1)/(L2) = (n1/n2).(300/750) = (2/3).(300/750) = 0.2667

But L1 + L2 = 50

thus, L1 + L1/0.2667 = 50

L1 = 10.53 cm