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Question Number 11969 by tawa last updated on 08/Apr/17
A gas occupies 30 dm^3  at s t p,  what volume will it occupy at 91°C   and 380 mmHg
$$\mathrm{A}\:\mathrm{gas}\:\mathrm{occupies}\:\mathrm{30}\:\mathrm{dm}^{\mathrm{3}} \:\mathrm{at}\:\mathrm{s}\:\mathrm{t}\:\mathrm{p},\:\:\mathrm{what}\:\mathrm{volume}\:\mathrm{will}\:\mathrm{it}\:\mathrm{occupy}\:\mathrm{at}\:\mathrm{91}°\mathrm{C}\: \\ $$$$\mathrm{and}\:\mathrm{380}\:\mathrm{mmHg} \\ $$
Answered by sandy_suhendra last updated on 08/Apr/17
V_1 =30 dm^3   P_1_  =1 atm  T_1 =273 K  P_2 =380 mmHg=((380)/(760)) atm=0.5 atm  T_2 =91+273=364 K  ((P_1 .V_1 )/T_1 )=((P_2 .V_2 )/T_2 )  ((1×30)/(273))=((0.5×V_2 )/(364))  V_2 =80 dm^3
$$\mathrm{V}_{\mathrm{1}} =\mathrm{30}\:\mathrm{dm}^{\mathrm{3}} \\ $$$$\mathrm{P}_{\mathrm{1}_{} } =\mathrm{1}\:\mathrm{atm} \\ $$$$\mathrm{T}_{\mathrm{1}} =\mathrm{273}\:\mathrm{K} \\ $$$$\mathrm{P}_{\mathrm{2}} =\mathrm{380}\:\mathrm{mmHg}=\frac{\mathrm{380}}{\mathrm{760}}\:\mathrm{atm}=\mathrm{0}.\mathrm{5}\:\mathrm{atm} \\ $$$$\mathrm{T}_{\mathrm{2}} =\mathrm{91}+\mathrm{273}=\mathrm{364}\:\mathrm{K} \\ $$$$\frac{\mathrm{P}_{\mathrm{1}} .\mathrm{V}_{\mathrm{1}} }{\mathrm{T}_{\mathrm{1}} }=\frac{\mathrm{P}_{\mathrm{2}} .\mathrm{V}_{\mathrm{2}} }{\mathrm{T}_{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}×\mathrm{30}}{\mathrm{273}}=\frac{\mathrm{0}.\mathrm{5}×\mathrm{V}_{\mathrm{2}} }{\mathrm{364}} \\ $$$$\mathrm{V}_{\mathrm{2}} =\mathrm{80}\:\mathrm{dm}^{\mathrm{3}} \\ $$
Commented by tawa last updated on 08/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 08/Apr/17
V=((nRT)/P)  so  V_2 = V_1 ((T_2 /T_1 ))((P_1 /P_2 ))  V_2 = 30 dm^3 (((364K)/(273K)))(((760mmHg)/(380mm Hg)))      = (30 dm^3 )((4/3))(2) = 80 dm^3  .
$${V}=\frac{{nRT}}{{P}}\:\:{so}\:\:{V}_{\mathrm{2}} =\:{V}_{\mathrm{1}} \left(\frac{{T}_{\mathrm{2}} }{{T}_{\mathrm{1}} }\right)\left(\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right) \\ $$$${V}_{\mathrm{2}} =\:\mathrm{30}\:{dm}^{\mathrm{3}} \left(\frac{\mathrm{364}{K}}{\mathrm{273}{K}}\right)\left(\frac{\mathrm{760}{mmHg}}{\mathrm{380}{mm}\:{Hg}}\right) \\ $$$$\:\:\:\:=\:\left(\mathrm{30}\:{dm}^{\mathrm{3}} \right)\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\left(\mathrm{2}\right)\:=\:\mathrm{80}\:{dm}^{\mathrm{3}} \:. \\ $$
Commented by tawa last updated on 08/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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