A cylinder containing an ideal gas has a volume of 2.6 m3 and a pressure of 1.5 × 105 Pa at a temperature of 300 K. The cylinder is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K. The change in internal energy of the gas is +6.0 × 105 J. How much heat did the gas absorb?

Respuesta :

Answer:[tex]13.5\times 10^{8}[/tex] joules

Explanation:

From the first law of thermodynamics,

Δ[tex]Q[/tex]=Δ[tex]U[/tex]+[tex]W[/tex]

Where [tex]Q[/tex] is the heat given to the gas,

[tex]U[/tex] is the internal energy of the gas,

[tex]W[/tex] is the workdone by the gas.

When pressure is constant,

[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]

[tex]V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}[/tex]

When pressure is constant,[tex]W=P[/tex]Δ[tex]V[/tex]

Where [tex]P[/tex] is pressure and [tex]V[/tex] is the volume of the gas.

Given [tex]P=1.5\times 10^{5}Pa[/tex]

Δ[tex]V=[/tex][tex]7.8-2.6=5.2m^{3}[/tex]

So,[tex]W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J[/tex]

Given that Δ[tex]U=6\times 10^{5}[/tex]

So,Δ[tex]Q=[/tex][tex]6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J[/tex]

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