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blsea [12.9K]
2 years ago
10

Can somebody please help me ! :(

Chemistry
2 answers:
avanturin [10]2 years ago
8 0

Answer:

325

Explanation:

Helen [10]2 years ago
5 0

Answer:

325

Explanation:

velocity = 3250 Hz x 0.1 m

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A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of
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<u>Answer:</u> The volume when the pressure and temperature has changed is 1.6\times 10^2mL

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

Let us assume:

P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K

Putting values in above equation, we get:

\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL

Hence, the volume when the pressure and temperature has changed is 1.6\times 10^2mL

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