Answer:
The new temperature of the gas is 604K
Explanation:
Assuming standard temperature and pressure, to calculate the temperature of the gas we use the general gas equation;
Step 1 :write the general gas equation
P1V1/ T1 = P2V2/T2
Step 2: Write out the values, covert the necessary values to the standard values.
P1 = 0.800atm.
V1 = 0.180L
T1 = 29°C = 273 + 29 = 302K
P2 , 3.20atm
V2 = 90mL = 90 * 10^-3L = 0.09L
Step 3: Solve for T2
The new temperature T2 of the gas is:
T2 = P2V2T1/ P1V1
T2 = 3.20 * 0.09 * 302 / 0.800 * 0.180
T2 = 86.976 / 0.144
T2 = 604K
The new temperature is of the gas is 604K.
A sample of helium gas occupies 0.180 L at 0.800 atm and 29 °C. It occupies 90 mL at 3.20 atm and 331 °C.
A sample of helium gas has a volume of 0.180L (V₁), a pressure of 0.800 atm (P₁) and a temperature of 29 °C (T₁).
We will convert 29 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 29\° C + 273.15 = 302K[/tex]
We want to calculate the temperature (T₂) at which it has a volume of 90 mL (V₂) and a pressure of 3.20 atm (P₂). We will use the combined gas law.
[tex]\frac{T_1}{P_1 \times V_1 } = \frac{T_2}{P_2 \times V_2 } \\\\T_2 = \frac{T_1 \times P_2 \times V_2}{P_1 \times V_1 } = \frac{302K \times 3.20atm \times 90mL}{0.800atm \times 180mL } = 604 K[/tex]
We can convert 604 K to Celsius.
[tex]\° C = K - 273.15 = 604-273.15 = 331 \° C[/tex]
A sample of helium gas occupies 0.180 L at 0.800 atm and 29 °C. It occupies 90 mL at 3.20 atm and 331 °C.
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