Question

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 $2500c{m}^3$.

The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal.

• A. 30.6 J, 1.94 bar
• B. 21.8 J, 0.93 bar
• C. 61.1 J, 1.94 bar
• D. 43.7 J, 0.93 bar

Answer 1

The correct option is D 43.7 J, 0.93 bar

Given data:
Gauge pressure,
$p_{G1}=1bar=100kPa$
Ambient temperature,
$T_1={15}^{o}C=(15+273)K=288K$
Absolute pressure,
$p_1=p_{G1}+p_{atm}$
$=100+101.325$
$=201.325kPa$
Volume: V = 2500 c$m^3$ = 2500 $\times {10}^{-6}{m}^3$
Temperature,
$T_2=5^{o}C=(5+273)K=278K$
Methode I:
From equation of state,
$p_1{V}_1=mR{T}_1$
$201.325\times 2500\times {10}^{-6}=m\times 0.287\times 288$
$m=6.089\times {10}^{-3}kg$
Heat loss: Q = $m{c}_v(T_1-T_2)$
= 6.089 $\times$ ${10}^{-3}$ $\times$ 0.718 $\times$ $\times$(288 - 278)
= 0.0437 kJ = 43.7 J
At V = C
$\frac{p_1}{T_1}=\frac{p_2}{T_2}$
$\frac{201.325}{288}=\frac{p_2}{278}$
Or $p_2=194.33kPa$ (absolute)
also$p_2=p_{G2}+p_{atm}$
$\therefore$ 194.33 = $p_{G2}$ + 101.325
Or $p_{G2}$ = 93 kPa = 0.93 bar

Method II:
$p_{2}V=mR{T}_2$
$p_2$ $\times$ 2500 $\times$${10}^{-6}$ = 6.089$\times$${10}^{-3}$$\times$ 0.287
Or $p_2$ = 194.32 kPa (absolute)
also $p_2$ = $p_{G2}$ + $p_{atm}$
$\therefore$ 194.32 = $p_{G2}$ + 101.325
Or $p_{G2}$ = 92.99 kPa = 93 kPa = 0.93 bar