Answer:
The value is [tex]\Delta T = 62.2 \ K[/tex]
Explanation:
From the question we are told that
The number of students is [tex]N = 138[/tex]
The metabolic rate for each student is [tex] r = 128 W[/tex]
The time duration is [tex]t = 60 \ minutes = 3600 \ s[/tex]
The molar specific heat of air is [tex]C_V =\frac{ 5}{2} R[/tex]
The volume is [tex]V = 1190 \ m^3[/tex]
The pressure is [tex]P = 1.01 *10^{5} \ Pa[/tex]
The initial temperature is [tex]T_i = 21^o C = 294[/tex]
Generally the metabolic rate of the students is
[tex] K = N * r [/tex]
=> [tex]K = 138 *128[/tex]
=> [tex]K = 17664 \ W[/tex]
The total heat generated by the students is
[tex]H = K * t[/tex]
=> [tex]H = 17664 * 3600[/tex]
=> [tex]H =6.3590*10^{7} \ J [/tex]
From the ideal gas law we can evaluate n (number of moles ) as
[tex]n = \frac{PV}{RT_i}[/tex]
Here R is the gas constant with value [tex]R = 8.314 \ J / mol . K[/tex]
So
[tex]n = \frac{1.01 *10^{5} * 1190 }{8.314 * 294}[/tex]
=> [tex]n = 49171.2 \ moles [/tex]
Generally the heat generated is mathematically represented as
[tex]H = n * C_V * \Delta T[/tex]
=> [tex]H = n * (\frac{ 5}{2} *R)* \Delta T[/tex]
=> [tex]\Delta T = \frac{2 * H}{n * 5 *R }[/tex]
=> [tex]\Delta T = \frac{2 * 6.3590*10^{7}}{49171.2 * 5 * 8.314}[/tex]
=> [tex]\Delta T = 62.2 \ K[/tex]
