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Question Number 16079 by Tinkutara last updated on 17/Jun/17

An open flask contains air at 27°C. To  what temperature it must be heated to  expel one-fourth of the air?

$$\mathrm{An}\:\mathrm{open}\:\mathrm{flask}\:\mathrm{contains}\:\mathrm{air}\:\mathrm{at}\:\mathrm{27}°\mathrm{C}.\:\mathrm{To} \\ $$$$\mathrm{what}\:\mathrm{temperature}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathrm{heated}\:\mathrm{to} \\ $$$$\mathrm{expel}\:\mathrm{one}-\mathrm{fourth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{air}? \\ $$

Commented by ajfour last updated on 18/Jun/17

is it 127°C ?  nR△T+(△n)RT =P△V+V△P   n△T+T△n =0   △T =((−T(△n))/n)           = −T(−(1/4)) =(T/4)   T−T_0 = (T_0 /4)     ⇒  T= 1.25T_0    but what is normal temperature   T= 25°   or  T=0°  ?

$${is}\:{it}\:\mathrm{127}°{C}\:? \\ $$$${nR}\bigtriangleup{T}+\left(\bigtriangleup{n}\right){RT}\:={P}\bigtriangleup{V}+{V}\bigtriangleup{P} \\ $$$$\:{n}\bigtriangleup{T}+{T}\bigtriangleup{n}\:=\mathrm{0} \\ $$$$\:\bigtriangleup{T}\:=\frac{−{T}\left(\bigtriangleup{n}\right)}{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:−{T}\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\frac{{T}}{\mathrm{4}} \\ $$$$\:{T}−{T}_{\mathrm{0}} =\:\frac{{T}_{\mathrm{0}} }{\mathrm{4}}\:\:\:\:\:\Rightarrow\:\:{T}=\:\mathrm{1}.\mathrm{25}{T}_{\mathrm{0}} \\ $$$$\:{but}\:{what}\:{is}\:{normal}\:{temperature} \\ $$$$\:{T}=\:\mathrm{25}°\:\:\:{or}\:\:{T}=\mathrm{0}°\:\:? \\ $$$$ \\ $$

Answered by mrW1 last updated on 18/Jun/17

T_2 /T_1 =V_2 /V_1 =1.25  T_2 =1.25T_1   t_2 =T_2 −273=1.25(27+273)−273=102°C

$$\mathrm{T}_{\mathrm{2}} /\mathrm{T}_{\mathrm{1}} =\mathrm{V}_{\mathrm{2}} /\mathrm{V}_{\mathrm{1}} =\mathrm{1}.\mathrm{25} \\ $$$$\mathrm{T}_{\mathrm{2}} =\mathrm{1}.\mathrm{25T}_{\mathrm{1}} \\ $$$$\mathrm{t}_{\mathrm{2}} =\mathrm{T}_{\mathrm{2}} −\mathrm{273}=\mathrm{1}.\mathrm{25}\left(\mathrm{27}+\mathrm{273}\right)−\mathrm{273}=\mathrm{102}°\mathrm{C} \\ $$

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