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Answer 1

EXPLANATION.

An ideal gas has a specific heat at a

constant pressure = Cp = (5/2) R

The gas kept in a close vessels of volume

=> 0.0083 m³.

=> Temperature = 300 k

=> pressure = 1.6 X 10^6 Nm^-2

=> an amount of 2.49 X 10^4 j of heat energy

is supplied to the gas.

To find the final temperature and

pressure of the gas.

According to the question,

$\rm \to \: c_{v} \: = \: c_{p} \: - \: r \\ \\ \rm \to \: c_{v} \: = \frac{5}{2} r \: - r \\ \\ \rm \to \: c_{v} \: = \frac{3r}{2} = 1.5r \\ \\ \rm \to \: as \: we \: know \: that \\ \\ \rm \to \: pv \: = nrt \\ \\ \rm \to \: n \: = \frac{pv}{rt} \\ \\ \rm \to \: put \: the \: value \: in \: equation \\ \\ \rm \to \: n \: = \frac{(1.6 \times 10 {}^{6}nm {}^{ - 2} ) \times \: (0.0083m {}^{3} )}{(8.314 \: j \: k {}^{ - 1}mol {}^{ - 1} ) \times (300k)} \\ \\ \rm \to \: n \: = 5.33 \: mol \\ \\ \rm \to \: as the \: gas \: kept \: in \: the \: vessels \\ \\ \rm \to \: \Delta \: Q \: = n c_{v}\Delta \: t \\ \\ \rm \to \: \Delta \: t \: = \frac{\Delta \: Q}{n c_{v} } \\ \\ \rm \to \: \frac{(2.49 \times 10 {}^{4}j) }{(5.33 \: mol) \times (1.5 \: \times 8.314 \: j \: k {}^{ - 1} mol {}^{ - 1}) } = 377 \\ \\ \rm \to \: \Delta \: t \: = 300k \\ \\ \rm \to \: final \: temperature \: = 300 + 377 = 667k \\ \\ \rm \to \: as \: we \: know \: that \\ \\ \rm \to \: \frac{ p_{1} v_{1} }{ t_{1}} = \frac{ p_{2} v_{2}}{ t_{2}} \\ \\ \rm \to \: volume \: is \: equal \: \\ \\ \rm \to \: v_{1} \: = \: v_{2} \\ \\ \rm \to \: p_{2} \: = \: \frac{ t_{2}p_{1} }{ t_{1} } \\ \\ \rm \to \: p_{2} \: = \frac{677}{300} \times 1.6 \times 10 {}^{6} nm {}^{ - 2} \\ \\ \rm \to \: p_{2} \: = 3.6 \times 10 {}^{6} nm {}^{ - 2}$

Answer 2

Answer:

$v{}{1 = v \frac{}{2} }$

$p2 = \frac{t2p1}{t1}$

$p2 = 3.6 \times 10 {}^{6} nm {}^{ - 2}$