Be sure to answer all parts. What is the effect of each of the following on the volume of 1 mol of an ideal gas? (a) The pressure changes from 760 torr to 202 kPa, and the temperature changes from 37°C to 155 K. The volume increases by a factor of 2. The volume increases by a factor of 4. The volume decreases by a factor of 2. The volume decreases by a factor of 4. The volume remains the same. (b) The temperature changes from 305 K to 32°C, and the pressure changes from 2 atm to 101 kPa. The volume increases by a factor of 2. The volume increases by a factor of 4. The volume decreases by a factor of 2. The volume decreases by a factor of 4. The volume remains the same.

Answer

A)The volume decreases by a factor of 4

B), the volume has increased by factor of 2

Explanation:

A)Given:

P1= 760Kpa

P2 =202Kpa

The temperature changes from37C to155C

There is increase In pressure from P1 to P2

P1= 760torr.

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

We need to convert the temperature from Celsius to Kelvin

T1= 37+273= 310K

But from ideal gas, we know that PV = nRT where nR is constant

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)*(310/155)

V1/V2=4

Therefore, the volume has decreased by factor of 4

B)

Given:

P1= 2atm

P2 =101Kpa

The temperature changes from 305K to 32C

There is increase In pressure from P1 to P2

P1= 2atm

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

P1= 202.65kpa

We need to convert the temperature from Celsius to Kelvin

T2= 32+273= 305K

But from ideal gas, we know that PV = nRT

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)

V1/V2 = (101/202.65)*(305/305)

V1/V2 = 1/2

Therefore, the volume has increased by factor of 2

So the empirical formula is C14H18N2O5

Explanation:

C = 57.2% = 12g/mol

H = 6.1% = 1g/mol

N = 9.5% = 14g/mol

O = 27.2% = 16g/mol

Empirical Formula for compound hmm

Assume

C = 57.2g

H = 6.1g

N = 9.5g

O = 27.2g

So we have

C = 57.2g/12g =  4.76  moles

H = 6.1g/1g      =  6.10   moles

N = 9.5g/14g   =  0.68  moles

O = 27.2g/16g =  1.70   moles

Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number.

C =  4.76  moles / 0.68 moles = 7

H = 6.10   moles / 0.68 moles = 9

N = 0.68  moles / 0.68 moles = 1

O = 1.70   moles / 0.68 moles = 2.5

Ok so we now have the ratios but for O it's 2.5, have to be whole numbers so we will need to double all the numbers.

C = 14

H = 18

N =  2

O =  5

So the empirical formula is C14H18N2O5