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Mechanical Properties Of Fluids

Question
CBSEENPH11020339

The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the numbers of molecules in the room before and after heating, then nf – ni will be

  • 2.5 x 1025

  • -2.5 x 1025

  • -1.61 x 1023

  • 1.38 x 1023

Solution

B.

-2.5 x 1025

Using ideal gas equation
PV = nRT
(N is number of moles)
P0V0 = niR × 290 ...... (1)
[Ti = 273 + 17 = 290 K]
After heating
P0V0 = NfR × 300 ...... (2)
[Tf = 273 + 27 = 300 K]
from equation (1) and (2)
straight N subscript straight f minus straight N subscript straight i space equals space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript 0 over denominator straight R space straight x 290 end fraction
difference space in space number space of space moles space minus fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 290 end fraction
difference space in space number space of space moles space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets
Hence comma space straight n subscript straight f space minus straight n subscript straight i space is
equals space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets straight x space 6.023 space straight x space 10 to the power of 23

putting P0 = 105 PA and V0 = 30 m3
Number of molecules nf – ni = – 2.5 × 1025