Question

Under constant pressure, the temperature of 2.43 mol of an ideal monatomic gas is raised 11.9 K. What are (a) the work W done by the gas, (b) the energy transferred as heat Q, (c) the change ΔEint in the internal energy of the gas, and (d) the change ΔK in the average kinetic energy per atom

Answer 1

Answer:

Explanation:

Given

no of moles $n=2.43$

Temperature raised $\Delta T=11.9 k$

Work done by gas

$W=\int_{V_1}^{V_2}PdV$

$W=P\Delta V$

$W=nR\Delta T$

$W=2.43\times 8.314\times 11.9$

$W=240.41 kJ$

(b)Energy Transferred as heat

$Q=nc_p\Delta T$

$c_p$=specific heat at constant Pressure

$c_p$ for ideal Mono atomic gas is $\frac{5R}{2}$

$Q=2.43\times \frac{5R}{2}\times 11.9$

$Q=601.03 kJ$

(c)Change in Internal Energy

$\Delta U=Q-W$

$\Delta U=601.03-240.41=360.62 kJ$

(d)Change in average kinetic Energy $\Delta k$

$K.E._{avg}=\frac{3}{2} \times k\times T$

$\Delta K.E.=\frac{3}{2} \times k\times \Delta T$  ,where k=boltzmann constant

$\Delta K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 11.9$

$\Delta K.E.=2.46\times 10^{-22} J$