Question

Part A

Calculate the volume of the gas, in liters, if 1.60 mol has a pressure of 1.20 atm at a temperature of -5 ∘C.

Express the volume in liters to three significant digits.

V V =	nothing	L

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Part B

Calculate the absolute temperature of the gas at which 3.53×10⁻³ mol occupies 478 mL at 760 torr .

Express the temperature in kelvins to three significant digits.

T T =	nothing	K

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Part C

Calculate the pressure, in atmospheres, if 2.65×10⁻³ mol occupies 415 mL at 135 ∘C.

Express the pressure in atmospheres to three significant digits.

P P =	nothing	atm

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Part D

Calculate the quantity of gas, in moles, if 126 L at 60.0 ∘C has a pressure of 11.15 kPa .

Express the amount in moles to three significant digits.

n n =	nothing	mol

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Answer 1

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(Pont A n=1-60mo, ofere we we ideal gas equation (PV=nRT p = pressure ; na no. of moles ; T = Temperature (k) V= volume ; R = Gas constant : = 0.0821 Liam.mock! n=1.60mol, p=1.20 ates, T= -5c+273 = 268k ; V=?? → V= nRT (1.6m6) (0.0821 Leatsh.matik) (268K) 1-2 alam V=35.20 L = 29.3 L Port B) = 3.53x103 mol , v=478ml = 0.478L (1000ml = 14) P = 760 torx = laten (in atm = 760 for), T=? > T = PV = (athm (0.4784) NR (0.0821 K. atm. mst! I') (3.53 x 103 m) T= 0.478K = 0.478 K = 16 48 1650 (0.0821 (0.00353) 0.00029 (3 Sight cant figura Past c n = 2.65x10° mof; v= 415 mL = 0.4152 T=135°C +273 = 408 k ; P= ? as P= nRT - (2.65x10 móU (0.0821 kata-mat. 09 (408 KS) 0.4154 p= 0.0887 atm = 10.214 aten 0.415 Post - 62? v= 126L & T = 60.0°C + 273=333k P = 11.15KPa Ikpa = 0.00987 atm 11.15kla= a alm. 11.15 kPa x0.00987 alim – 0-110aton (kPa 1386 mol na par - (0.11 afr) (1264) (0.0811 Katsch. month (333) = 27. The 0.506 mol