Question

What is the temperature of 0.750 mol of a gas stored in a 6,850 mL cylinder at 2.21 atm? A: 2.95 K

B:5.24 K
C:138 K
 D:246 K

Answer 1

D is the correct answer

Answer 2

Option D: 246 K

According to ideal gas law:

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

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature of the gas. The value of R is
`0.082 atm L K^(-1)mol^(-1)`.

Number of moles of gas are 0.750 mol, volume of cylinder is 6850 mL and pressure of gas is 2.21 atm.

First converting volume from mL to L

`1 mL=10^(-3)L`

Thus,

6850 mL=6.850 L

Putting the values in equation (1),

`(2.21 atm)(6.850 L)=(0.750 mol)(0.082 atm L K^(-1)mol^(-1)T`

Rearranging

`T=((2.21 atm)(6.850 L))/((0.750 mol)(0.082 atm L K^(-1)mol^(-1)))=246 K`,

Therefore, temperature of the gas is 246 K.