SSC JE Mechanical Engineering 2017 Set 3 Paper

Question : 101 of 200

Marks: +1, -0

MECHANICAL

A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 15°C.When the game started next day, the air temperature at the stadium was 5°C. Assume that the volume of the football remains constant at 2500

cm^3

. Gauge pressure of air to which the ball must have been originally inflated so that it would equal 1 bar gauge at the stadium is______.

2.23 bar
1.94 bar
1.07 bar
1 bar

Validate

Solution:

P_1=100+101.325

= 201.325kPa

T_1=288K,V_1=2500\cm^3

m={P_1V_1}/{RT_1}

={201.325×2500×10^{-6}}/2

=6.089×10^{-3}\kg

as dv = 0 ⇒ w = 0
dQ = dU = mqdT

=6.089×10^{-3}×0.718×10^3(15-5)

dQ = 43.75

P_1/T_1=P_2/T_2

P_2=P_1×T_2/T_1

={201.325×278}/288

= 194.33kpa (absolute)
Gauge pressure on day of play = 194.33 – 101.325 = 93kPa = 0.93 bar.

P_1

= Pressure to which it must be inflated

P_1={201.325×288}/278

= 208.56 kPa.
Gauge pressure to which it must be inflated on previous day
= 208.56 – 101.325 kPa
= 107.24kPa = 1.07 bar

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