Question

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 1

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

$V2= V1/4$

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

$V2=2V1$

Therefore, the volume has increased by factor of 2

Answer 2

A)The volume has decreased by a factor of 4.

B) The volume has increased by factor of 2.

Given:

P₁ = 760 KPa

P₂ =202KPa

The temperature changes from 37°C to 155 K

P₁ = 760torr.

Ideal gas equation:

PV = nRT ..............(1)

where;

nR is constant

P= pressure

V= volume

T= temperature

n = number of moles

On adding values:

(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 =101 KPa

The temperature changes from 305 K to 32°C

P₁= 2atm

On adding values in equation (1)

(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.

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