Answer:
[tex]T_2=158.9K=-114.25^oC[/tex]
Explanation:
Hello,
In this case, we use the combined gas law which allows us to understand the pressure-volume-temperature behavior as shown below:
[tex]\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}[/tex]
In such case, solving the temperature at the end (in kelvins and degrees Celsius) we obtain:
[tex]T_2= \frac{P_2V_2T_1}{P_1V_1}=\frac{222kPa*1.03L*(-17+273.15)K}{1.94L*190kPa} \\\\T_2=158.9K=-114.25^oC[/tex]
Best regards.
Answer:
The new temperature will be 158.8 K
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
