Question

please show me how to get the correct answer in preperation for my upcoming exam

1.) B 3.) D 4.) C

Chemistry 1171 Practice Exam 3 (Chapter 5 & 6) 1. If 30.0 L of oxygen are cooled at constant pressure from 200°C to 1°C, what is the new volume of oxygen? A. 0.150 L B . 17.4 L c . 23.0L D. 51.8 L E. 6.00 x 10 L

Answer 1

Ans:1

We know, according to the ideal gas equation:

PV = nRT

Where P = pressure

V = volume

R = gas constant

T = temperature in K

so, Now, according to question (At constant P)

PV1 = nRT1 ..................(1)

PV2 = nRT2 ..................(2)

diving equation (1) by (2), we get

PV1/PV2 = nRT1/nRT2

V1/V2 = T1/T2

30.0L / V2 = 473K / 278K

V2 = 30.0L * 278 / 473

V2 = 17.4L

Option B is correct answer.

Ans:2

According to the ideal gas equation:

P1V1 = nRT1 ........... (3)

P2V2 = nRT2 ............(4)

Dividing equation (3) by (4), we get,

P1V1/ P2V2 = nRT1 / nRT2

P1V1/ P2V2 = T1 / T2 (as the amount of gas is fixed)

Now, putting the values we get,

760mmHg * 175L / 640mmHg * 198L = 288K / T2 (15 ^oC + 273 = 288K)

T2 = 288K * 640 *198 / 760 * 175

= 274.4K

We know,

the temperature in Kelvin = the temperature in degree Celcius + 273

so, the temperature in degree Celcius = the temperature in Kelvin - 273

= (274 - 273) ^oC

=1 ^oC

hence, option D is the answer.

Ans:3

According to the ideal gas equation:

PV =nRT

n = weight of CO₂ / molar mass of CO₂

= 25.2g / 44g

= 0.57moles

Value of R in atm = 0.082057 L atm mol^-1K^-1

Temperature in K = (25+273 )K = 298K

Now, putting values in ideal gas equation,

0.840atm *V = 0.57moles * 0.082057 L atm mol^-1K^-1 * 298K

V = 16.9L

hence, option C is correct.

Hope this helps!