Question

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?

Answer 1

Answer:$13.5\times 10^{8}$ joules

Explanation:

From the first law of thermodynamics,

Δ$Q$=Δ$U$+$W$

Where $Q$ is the heat given to the gas,

$U$ is the internal energy of the gas,

$W$ is the workdone by the gas.

When pressure is constant,

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

$V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}$

When pressure is constant,$W=P$Δ$V$

Where $P$ is pressure and $V$ is the volume of the gas.

Given $P=1.5\times 10^{5}Pa$

Δ$V=$$7.8-2.6=5.2m^{3}$

So,$W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J$

Given that Δ$U=6\times 10^{5}$

So,Δ$Q=$$6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J$