Question #9c03e

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Answer 1 · 2016-06-23T07:57:20

9.72atm

Given

$V_{1} \to Initial volume of air = 400 c m^{3}$ $V_1 ->"Initial volume of air"=400cm^3$

$P_{1} \to Initial pressure air = 1 a t m$ $P_1 ->"Initial pressure air"= 1atm$

$T_{1} \to Initial temperature air = 15 + 273 = 288 K$ $T_1 ->"Initial temperature air"= 15+273=288K$

$V_{2} \to Final compressed volume of air = 50 c m^{3}$ $V_2 ->"Final compressed volume of air"=50cm^3$

$T_{2} \to Final temperature air = 77 + 273 = 350 K$ $T_2 ->"Final temperature air"= 77+273=350K$

$P_{2} \to Initial pressure air = ?$ $P_2 ->"Initial pressure air"=?$

Now we have combined Boyle's and Charle's law relation as

$\frac{P_{2} V_{2}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$ $(P_2V_2)/T_2=(P_1V_1)/T_1$

$\Rightarrow \frac{P_{2} \cdot 50}{350} = \frac{1 \cdot 400}{288}$ $=>(P_2*50)/350=(1*400)/288$

$P_{2} = \frac{8 \cdot 350}{288} = 9.72 a t m$ $P_2=(8*350)/288=9.72atm$