Question

The temperature of 2.0 g of helium is increased at constant volume by ΔT. What mass of oxygen can have its temperature increased by the same amount at constant volume using the same amount of heat?

Answer 1

Answer:

m = 9.6 g

Explanation:

Thermal energy given to helium gas at constant volume is given as

$Q = nC_v \Delta T$

so here we have

$C_v = \frac{3}{2}R$

$n = moles$

$n = \frac{2}{4} = 0.5$

so we have

$Q = \frac{3}{2}R(0.5)\Delta T$

now we know that

for oxygen gas we have

$C_v = \frac{5}{2}R$

for same amount of heat we have

$Q = nC_v \Delta T'$

$\frac{3}{2}R(0.5)\Delta T = \frac{m}{32} (\frac{5R}{2}) \Delta T$

$m = \frac{0.75 \times 32}{2.5}$

$m = 9.6 g$