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3 years ago

5

The volume of a gas is the same as its _____.

2 answers:

AleksandrR [38]3 years ago

8 0

The volume of a gas is the same as its CONTAINER.

Gases generally has no shape and no definite volume. When a gas is placed in a container, the gas usually takes the shape and the volume of the container, that is, the gas fills up all the available spaces in the container. Thus, the volume of a gas will always be the same as its container. This is in contrast with solids, which have definite shape and volume and liquids, which have definite volume but no fixed shape.

rodikova [14]3 years ago

6 0

My best guess is mass

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3 years ago

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

arsen [322]

Answer:

a) $T_b=590.775k$

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

d) $Q=3.778*10^4J$

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

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

$t = 20+273$

$t = 293k$

Initial heat $Q=1.36 * 10^4 J$

a)

Generally the equation for change in temperature is mathematically given by

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

Where

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

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

$T_b-293k=297.775$

$T_b=590.775k$

b)

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

$PV=Wbc$

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

$Wbc=10805.7J$

Total Work-done $W_t$

$W_t=Wab+Wbc$

$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

c)

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

$Q=nC_p\triangle T$

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

$Q=3.778*10^4J$

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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Answer:

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