Question

A gas has a volume of 1.94 L at −17◦C and

190 kPa. At what temperature would the gas

occupy 1.03 L at 222 kPa?

Answer 1

The new temperature will be 158.8 K

Step-by-step explanation:

Step 1: Data given

Volume of the gas = 1.94 L

The temperature = -17°C = 256 K

Pressure = 190 kPa

The volume reduces to 1.03 L

The pressure increases to 222 kPa

Step 2: Calculate the temperature =

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 = the initial pressure = 190 kPa

⇒with V1 = the initial volume = 1.94 L

⇒with T1 = -17°C = 256 K

⇒with P2 = the increased pressure = 222 kPa

⇒with V2 = the reduced volume = 1.03 L

⇒with T2 = the new volume

(190 kPa* 1.94 L) / 256 K = (222 kPa *1.03 L) T2

1.439844 = (222 * 1.03 )/T2

1.439844 = 228.66 / T2

T2 = 228.66 / 1.439844

T2 = 158.8 K

The new temperature will be 158.8 K

Answer 2

`T_2=158.9K=-114.25^oC`

Step-by-step explanation:

Hello,

In this case, we use the combined gas law which allows us to understand the pressure-volume-temperature behavior as shown below:

`(P_1V_1)/(T_1)= (P_2V_2)/(T_2)`

In such case, solving the temperature at the end (in kelvins and degrees Celsius) we obtain:

`T_2= (P_2V_2T_1)/(P_1V_1)=(222kPa*1.03L*(-17+273.15)K)/(1.94L*190kPa) \\\\T_2=158.9K=-114.25^oC`

Best regards.