5) A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature fr

Question

5) A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The 5) final pressure is atm. A) 15 B) 7.5 C) 3.3 D) 0.67 E) 1.5 6) If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy L at STP A) 3.92 B) 12.9 C) 25.5 D) 50.8 E) 5.08 7) A sample of a gas originally at 29 C and 1.25 atm pressure in a 3.0L container is allowed to contract until the volume is 2.2 L and the temperature is 11 °C The final pressure of the gas is atm. A) 0.38 B) 1.6 C) 2.9 D) 2.8 E) 21 mod. 8) The amount of gas that occupies 3652 L at 680 "C and 672 mm Hg is A) 127 B) 878 C) 12.7 D) 1.15 E) 24.4 °C 9) The temperature of a sample of CH4 gas (10.34 g) in a 500 L vessel at 1.33 atm is C) 195 D)-195 E)-1260 A) 984 B) 1260

Explanation / Answer

Ans 5) According to Boyle's Law; P1V1 = P2V2 (Fixed Amount and Constant Temperature)

Where, P1 = Initial Pressure (1 atm) V1 = Initial Volume = 10 L

P2 = Final Pressure V2 = Final Volume 15 L

So, 1 x 10 = P2 x 15

Hence P2 = 0.67 atm [D]

Ans 6) At STP, mass of a gas is directly preportional to its volume.

Therefore, 50.75 gm = 10 L

129.3 gm = ? L

So, 129.3 x 10 / 50.75

Ans = 25.5 gm [C]

Ans 7) Combine gas equation is P1V1/T1 = P2V2/T2

Where, P1 = Initial Pressure (1.25 atm) V1 = Initial Volume (3 L) T1 = Initial Temperature (29 + 273 = 302K)

P2 = Final Pressure V1 = Final Volume (2.2 L) T2 = Final Temperature (11 + 273 = 284K)

So, 1.25 x 3/302 = P2 x 2.2/ 284

Therefore P2 = 1.6 atm [B]

Ans 8) PV=nRT

Where, P = Pressure (672/760 atm);

V = Volume (36.52 L);

R = gas constant (0.082 atm L K-1 mol-1);

T = Temperature (68 + 273 = 341 K )

n = no of moles

n = (672 x 36.52)/(0.082 x 760 x 341)

So n =1.15 mol [D]

Ans 9) PV=nRT

Where, P = Pressure (1.33 atm);

V = Volume (50 L);

R = gas constant (0.082 atm L K-1 mol-1);

T = Temperature (? K )

n = no of moles [10.34/16 (mol wt of CH4]

T = (1.33 x 50 x 16)/ (10.34 x 0.082)

So T = 1255 K, Hence temperature in oC = 1255-273 = 982 oC

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