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Alenkinab [10]
3 years ago
10

The volume of a sample gas, initially at 25 C and 158 mL, increased to 450 mL. What is the final temperature of the sample of ga

s, if the pressure in the container it kept constant?
Chemistry
1 answer:
Rashid [163]3 years ago
3 0

Answer:

Final temperature of the gas is  576 ^{0}\textrm{C}.

Explanation:

As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:

                                       \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

where V_{1} and V_{2} are volume of gas at T_{1} and T_{2} temperature (in kelvin scale) respectively.

Here V_{1}=158mL , T_{1}=(273+25)K=298K and V_{2}=450mL

So  T_{2}=\frac{V_{2}T_{1}}{V_{1}}=\frac{(450mL)\times (298K)}{(158mL)}=849K 

849 K = (849-273) ^{0}\textrm{C} = 576 ^{0}\textrm{C}

So final temperature of the gas is  576 ^{0}\textrm{C}.

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A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the
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Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C

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Answer:

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Explanation:

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