2. Find the Volume of 3.00 mol of gas whose temperature is 60.0 C and whose pressure is 5.00 L. 3. Find the tempature of 1.50 mol of gas whose pressure is 4.00 atm and whose volume is 7.50L
4. Find the number of moles of a certain ideal gas whose volume, tempature and pressure are 3.00 L, 25.0, and 2.00 atm, respectively
Please show work I will give you a 5 star review and much points

2. 16.4 L

3. 243.6K

4. 0.25mol

Explanation:

Using the ideal gas formula for the series of questions in this case, we have;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

QUESTION 2:

V = ?

P = 5 atm

n = 3 mol

T = 60°C = 60 + 273 = 333K

PV = nRT

5 × V = 3 × 0.0821 × 333

5V = 82.02

V = 82.02/5

V = 16.4 L

QUESTION 3:

P = 4 atm

V = 7.5L

n = 1.5mol

T = ?

PV = nRT

4 × 7.5 = 1.5 × 0.0821 × T

30 = 0.123T

T = 30/0.123

T = 243.6K

QUESTION 4:

P = 2.00 atm

V = 3L

n = ?

T = 25°C = 25 + 273 = 298K

PV = nRT

2 × 3 = n × 0.0821 × 298

6 = 24.47n

n = 6/24.47

n = 0.25mol

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%