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Question Number 9982 by Tawakalitu ayo mi last updated on 20/Jan/17

Answered by sandy_suhendra last updated on 20/Jan/17

(1a) 1 atm=760 mmHg 30 mmHg=((30)/(760)) atm=0.0395 atm (1b) 1 torr=1 mmHg 30 mmHg=30 torr (1c) density of Hg=ρ=13,600 kg/m^3 gravity=g=9.8 m/s^2 30 mmHg=0.03 mHg=0.03×13,600×9.8=3,998.4 N/m^2 (1d) 1 N/m^2 =1 Pa 30 mmHg=3,998.4 Pa

$$\left(\mathrm{1a}\right)\:\mathrm{1}\:\mathrm{atm}=\mathrm{760}\:\mathrm{mmHg} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{30}\:\mathrm{mmHg}=\frac{\mathrm{30}}{\mathrm{760}}\:\mathrm{atm}=\mathrm{0}.\mathrm{0395}\:\mathrm{atm} \\ $$$$\left(\mathrm{1b}\right)\:\mathrm{1}\:\mathrm{torr}=\mathrm{1}\:\mathrm{mmHg} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{30}\:\mathrm{mmHg}=\mathrm{30}\:\mathrm{torr} \\ $$$$\left(\mathrm{1c}\right)\:\mathrm{density}\:\mathrm{of}\:\mathrm{Hg}=\rho=\mathrm{13},\mathrm{600}\:\mathrm{kg}/\mathrm{m}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{gravity}=\mathrm{g}=\mathrm{9}.\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{30}\:\mathrm{mmHg}=\mathrm{0}.\mathrm{03}\:\mathrm{mHg}=\mathrm{0}.\mathrm{03}×\mathrm{13},\mathrm{600}×\mathrm{9}.\mathrm{8}=\mathrm{3},\mathrm{998}.\mathrm{4}\:\mathrm{N}/\mathrm{m}^{\mathrm{2}} \\ $$$$\left(\mathrm{1d}\right)\:\mathrm{1}\:\mathrm{N}/\mathrm{m}^{\mathrm{2}} =\mathrm{1}\:\mathrm{Pa} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{30}\:\mathrm{mmHg}=\mathrm{3},\mathrm{998}.\mathrm{4}\:\mathrm{Pa} \\ $$

Answered by sandy_suhendra last updated on 20/Jan/17

(2) P_1 =p V_1 =v V_2 =5v P_1 V_1 = P_2 V_2 ⇒ p.v=P_2 .5v ⇒ P_2 =(1/5)p so the pressure will be (1/5) p [(1/5) times from the original] (3) T_1 =298 K P_1 =p V_1 =v P_2 =4p V_2 =4v ((P_1 V_1 )/T_1 ) = ((P_2 V_2 )/T_2 ) ⇒ ((p.v)/(298)) = ((4p.4v)/T_2 ) ⇒ T_2 =4,768 K

$$\left(\mathrm{2}\right)\:\mathrm{P}_{\mathrm{1}} =\mathrm{p} \\ $$$$\:\:\:\:\:\:\:\mathrm{V}_{\mathrm{1}} =\mathrm{v} \\ $$$$\:\:\:\:\:\:\:\mathrm{V}_{\mathrm{2}} =\mathrm{5v} \\ $$$$\:\:\:\:\:\:\:\mathrm{P}_{\mathrm{1}} \mathrm{V}_{\mathrm{1}} \:=\:\mathrm{P}_{\mathrm{2}} \mathrm{V}_{\mathrm{2}} \:\Rightarrow\:\mathrm{p}.\mathrm{v}=\mathrm{P}_{\mathrm{2}} .\mathrm{5v}\:\Rightarrow\:\mathrm{P}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{5}}\mathrm{p} \\ $$$$\:\:\:\:\:\:\:\mathrm{so}\:\mathrm{the}\:\mathrm{pressure}\:\mathrm{will}\:\mathrm{be}\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{p}\:\:\:\left[\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{times}\:\mathrm{from}\:\mathrm{the}\:\mathrm{original}\right] \\ $$$$\left(\mathrm{3}\right)\:\mathrm{T}_{\mathrm{1}} =\mathrm{298}\:\mathrm{K} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{P}_{\mathrm{1}} =\mathrm{p} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{V}_{\mathrm{1}} =\mathrm{v} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{P}_{\mathrm{2}} =\mathrm{4p} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{V}_{\mathrm{2}} =\mathrm{4v} \\ $$$$\frac{\mathrm{P}_{\mathrm{1}} \mathrm{V}_{\mathrm{1}} }{\mathrm{T}_{\mathrm{1}} }\:=\:\frac{\mathrm{P}_{\mathrm{2}} \mathrm{V}_{\mathrm{2}} }{\mathrm{T}_{\mathrm{2}} }\:\:\Rightarrow\:\frac{\mathrm{p}.\mathrm{v}}{\mathrm{298}}\:=\:\frac{\mathrm{4p}.\mathrm{4v}}{\mathrm{T}_{\mathrm{2}} }\:\Rightarrow\:\mathrm{T}_{\mathrm{2}} =\mathrm{4},\mathrm{768}\:\mathrm{K} \\ $$$$ \\ $$

Commented by Tawakalitu ayo mi last updated on 21/Jan/17

God bless you sir. i really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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