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3 years ago

9

A gas with a volume of 300 mL at 150 °C is heated until its volume is 600 mL. What is the new temperature of the gas if the pres

sure is unaltered?

1 answer:

Mandarinka [93]3 years ago

3 0

.012K

Use the equation of $\frac{V1}{T1}$ = $\frac{V2}{T2}$ .

Replace V with volume and T with the temperature. However, with the temperature, make sure to convert the celsius into kelvin by adding 273.

Hope this helps!

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A Snickers bar contains 27.0 g of sugars (carbohydrates), 4.0 g of protein and 12.0 g of fat. How many food calories (Cal) are c

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Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con

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b) $W_t=1.08*10^4J$

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Explanation:

From the question we are told that:

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Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

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a)

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

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$T_b=590.775k$

b)

Generally the equation for ideal gas is mathematically given by

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For v double

$T_c=2*590.775k$

$T_c=1181.55k$

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$Wbc=(2.20)(8.314)(1181_590.778)$

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Generally the equation for amount of heat added is mathematically given by

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$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

d)

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

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